Ahaan's trigonometric progression

By Infinite sum of gemetric series formula, since $0 \le \sin^2 \theta = (\sin \theta)^2 \le 1$ , then :

$a = \sin \theta$ , and $r = \sin^2 \theta$ , then,

$\sum_{n=0}^{\infty} \sin^{2n+1} \theta = \frac{a}{1 - r}$ $= \frac{\sin \theta}{1 - \sin^2 \theta} = 1 \ldots \ldots (1)$

Then, we obtain $\sin^2 \theta + \sin \theta - 1 = 0$ , where the solution is

$\sin \theta = \frac{\sqrt{5} - 1}{2} \implies \cos^2 \theta = 1 - sin^2 \theta = \frac{\sqrt{5} - 1}{2}$

Recall (1), again, $1 = \frac{\tan \theta}{\cos \theta} \implies \tan \theta = \cos \theta \implies 2 \tan^8 \theta = 2 (\cos^2 \theta)^4 = 7 - 3\sqrt{5}$

Then, $a + b + c = 7 + 3 + 5 = \boxed{15}$

We see that $\displaystyle\sum_{n=0}^{\infty} \sin^{2n+1} \theta = \sin^1 \theta + \sin^3 \theta + \sin^5 \theta + \cdots = \dfrac {\sin \theta}{1 - \sin^2 \theta} = 1.$ Here, we used the infinite sum of a geometric series. If we set $y = \sin \theta$ , we obtain $y = 1 - y^2$ . Now, we know that $\tan \theta = \dfrac {\sin \theta}{\sqrt {1 - \sin^2 \theta}} \implies \tan^8 \theta = \left( \dfrac {y^2}{1 - y^2} \right)^4.$ Now, we note that $\dfrac {y^2}{1 - y^2} = \dfrac {y}{1 - y^2} \cdot y = 1 \cdot y = y$ . Therefore, our answer is $2y^4$ . But recall that $y = 1 - y^2$ , so we solve: $y^2 + y - 1 = 0 \implies y = \dfrac {\sqrt{5}-1}{2},$ since $y$ is positive. We want $2y^4$ , so we expand and we eventually get $7 - 3 \sqrt {5}$ , so the answer is $7 + 3 + 5 = \boxed {15}$ .

The series says $\sin x + \sin^3 x + \sin^5 x....=1 \\ \implies \sin x + \sin^2 x ( \sin x + \sin^3 x + \sin^5 ) \\ \implies \sin x +\sin^2 x =1$ Therefore $\sin x = \frac{\sqrt {5} -1 }{2}$ $\\$ We also get $\sin^2 x = \cos x$ $$ Writing $\tan x$ as $\large \frac{sin x}{cos x}$ and then simplifying, we get $\tan^8 x = 7 - 3\sqrt{5}$ $$ So $\boxed{a+b+c = 15}$

$\Sigma$ $\sin^{2n+1} \theta$ = $\sin \theta$ (1 + $\sin^{2} \theta$ + $\sin^{4} \theta$ + ... ) = $\sin \theta$ ( $\frac{1}{1 - \sin^{2} \theta})$ = 1. Which gives $\sin \theta$ = $\cos^{2} \theta$ .Substituing on $sin^{2} \theta$ + $cos^{2} \theta$ = 1, give to us the following equation: $\cos^{4} \theta$ + $\cos^{2} \theta$ - 1=0, taking the positive root , we have: $\cos^{2} \theta$ = $\frac{\sqrt{5} - 1}{2}$ . We want 2 $\frac{ \sin^{8} \theta}{\cos^8 \theta}$ , but $\sin \theta$ = $\cos^{2} \theta$ , so we want

2 $\cos^{8} \theta$ = $\frac{(\sqrt{5} - 1)}{(2)^{4}}^{4}$ = 7 - 3 $\sqrt{5}$ . Hence 7 + 3 + 5 = 15

The given summation simply results in a geometric series, with first term $\sin\theta$ and common ratio $\sin^2\theta$ . The formula for the sum of an infinite geometric series is $\frac{a}{1-r}$ , so the given sum is equivalent to $\frac{\sin\theta}{1-\sin^2\theta}$ . Since this equals 1, we have that $\frac{\sin\theta}{1-\sin^2\theta}=1$ $\sin\theta=1-\sin^2\theta\ (1)$ $\sin\theta=\cos^2\theta\quad\ \ \ (2)$ Equation (2) follows from the Pythagorean identity $\sin^2\theta+\cos^2\theta=1$ .

Now, we must find $2\tan^8\theta=2\frac{\sin^8\theta}{\cos^8\theta}$ But by equation (2) $2\frac{\sin^8\theta}{\cos^8\theta}=2\frac{\sin^8\theta}{\sin^4\theta}=2\sin^4\theta$ All we have to do is find $\sin\theta$ , and we can, using the quadratic formula on equation (1). $\sin\theta=\frac{-1\pm\sqrt{5}}{2}$ Since $\theta$ is acute, $\sin\theta$ is positive. Finally, $2\left(\frac{\sqrt{5}-1}{2}\right)^4=7-3\sqrt{5}$ The answer is $7+3+5=\boxed{15}$

i) Evaluating the given summation ∑sin²ⁿ+¹(Ө) for n = 0 to ∞, This sum S = sin(Ө) + sin³Ө + sin⁵Ө + sin⁷Ө + ....... up to infinity

This is in GP, whose 1st term is sin(Ө) and common ratio is sin²Ө As Ө is an acute angle, sin(Ө) is in (0, 1); so, sin²Ө also in (0, 1) ==> |r| < 1 So the series sum forms infinite sum, which equal = a/(1 - r) = sinӨ/(1 - sin²Ө)

ii) Given sum = 1 ==> sinӨ = 1 - sin²Ө = cos²Ө ------ (1) Rearranging, sin²Ө + sinӨ - 1 = 0

This is a quadratic in sinӨ; solving sin(Ө) = (√5 - 1)/2 ----- (2) [Only positive values is taken; negative value is rejected, since Ө is acute]

iii) tan⁸Ө = sin⁸Ө/cos⁸Ө = sin⁸Ө/(cos²Ө)⁴ But from (1), cos²Ө = sinӨ

Substituting this in the above, it reduces to: tan⁸Ө = sin⁸Ө/sin⁴Ө = sin⁴Ө ----- (3)

iv) From (2), sin(Ө) = (√5 - 1)/2 ==> sin²Ө = [(√5 - 1)/2]² = (3 - √5 )/2

==> sin⁴Ө = [(3 - √5 )/2]² = (7 - 3√5 )/2

v) Hence of the above, 2 tan⁸Ө = 2 sin⁴Ө = 2*[ (7 - 3√5 )/2] = (7 - 3√5 )

==> a−b√c = 7 - 3√5; so a = 7; b = 3 and c = 5

Hence a + b + c = 15

Let $S=\sum_{n=0}^\infty \sin^{2n+1} \theta$

$S=\sin \theta + \sin^3 \theta + \sin^5 \theta \ldots$

Multiply both sides by $\sin^2 \theta$

$\sin^2\theta \cdot S=\sin^3 \theta+\sin^5 \theta \ldots$

Hence we get :

$S(1-\sin^2\theta)=\sin\theta$

Now since $S=1$ ,

$\Rightarrow \sin \theta= cos^2 \theta$

$\Rightarrow \sin^2 \theta+ \sin \theta -1=0$ )

As such we can re-write

$2 \tan^8 \theta$ as $2 sin^4 \theta$ .

Solving for $sin \theta$ in the quadratic gives us that $\large \sin \theta = \frac{-1 \pm \sqrt{5}}{2}$

Note that since $\sin \theta \geq -1$ we must have $\large \sin \theta =\frac{\sqrt{5}-1}{2}$

And expanding $2\sin^4 \theta$ gives us:

$7-3\sqrt{5}$

Hence our answer is : $\boxed{15}$

The given sum is clearly an infinite G.P. with common ratio $0$ $\leq$ $sin^{2}\theta$ $\leq1$ . Obviously $sin\theta$ cannot assume values at end points, because if $sin\theta$ = $0$ or $1$ the sum would be $0$ or $\infty$ respectively both of which do not agree with the given sum of $1$ . So $0<$ $sin^{2}\theta$ $<1$ .

Using the formula for sum of an infinite G.P. we get $sin\theta$ /(1- $sin^{2}\theta$ ) = $1$ -----( $\ast$ )

Solving eqn.( $\ast$ ) we get $sin\theta$ = $\frac{-1+\sqrt5}{2}$ -----( $\ast$ $\ast$ ) (rejecting the negative value since $\theta$ is an acute angle).

From eqn( $\ast$ ) we have $sin\theta$ = $cos^{2}\theta$ . So $tan^{2}\theta$ = $sin^{2}\theta$ / $cos^{2}\theta$ = $sin\theta$ . So $2tan^{8}\theta$ = $2$ ( $tan^{2}\theta$ )^4 = $2sin^{4}\theta$ = $7-3$ $\sqrt{5}$ = $a - b$ $\sqrt{c}$ . Comparing we obtain $a + b + c$ = $7 + 3 + 5$ = $\boxed{15}$ as our answer.

sinθ + sin^3θ + sin^5θ + sin^7θ +..........=1 _ (1) \Rightarrow sinθ(1 + sin^2θ + sin^4θ + sin^6θ +......)=1 \Rightarrow sinθ{ (sinθ + sin^3θ + sin^5θ + sin^7θ +..........) + sin^2θ + sin^4θ + sin^6θ +........ }=1 [By putting the value of '1', from the equation (1)] \Rightarrow (sinθ + sin^2θ + sin^3θ + sin^4θ + sin^5θ + sin^6θ + sin^7θ +..........)=\frac {1}{sinθ} __ (2) \Rightarrow sinθ( 1 + sinθ + sin^2θ + sin^3θ + sin^4θ + sin^5θ + sin^6θ + sin^7θ +..........)=\frac {1}{sinθ} \Rightarrow sinθ( 1 + \frac {1}{sinθ} )= \frac {1}{sinθ} [By putting the value of 'sinθ + sin^2θ + sin^3θ + sin^4θ + sin^5θ + sin^6θ + sin^7θ +..........' from the equation (2)] \Rightarrow sin^2θ + sinθ=1 \Rightarrow sin^2θ + sinθ - 1=0 \Rightarrow sinθ=\frac{-1 \pm \sqrt{1+4}}{2} We know that, θ is acute angel, so the negative value of sinθ must be eliminated. So, sinθ= \frac{-1 + \sqrt{1+4}}{2} \Rightarrow sin^2θ= \frac{3 - \sqrt{5}}{2} \Rightarrow cos^2θ=\frac{-1 + \sqrt{1+4}}{2} \Rightarrow tan^2θ=\frac{-1 + \sqrt{1+4}}{2} \Rightarrow tan^8θ=\frac{7 - 3\sqrt{5}}{2} \Rightarrow 2tan^8θ=\frac{7 - 3\sqrt{5}} So, a=7, b=3, c=5. Then, a+b+c=7+3+5=15[answer]

Let us denote the above sum by S.

S=sinθ +sin^3 θ +sin^5 θ......

S(sin^2 θ)=sin^3 θ+sin^5 θ+sin^7 θ.....

S-S(sin^θ)=sinθ

S(1-sin^2 θ)=sinθ

S=sinθ/1-sin^2 θ=1

from the above equation sinθ= (-1 ±√5)/2

also from the same equation sinθ=cos^2 θ

cos^2 θ=sin θ=(-1 ±√5)/2

now 2tan^8 θ=2sin^8 θ/cos^8 θ=2cos^8 θ=7-3√5=a-b√c

a+b+c=7+3+5=15

The given sum is an infinite G.P and as -1<=sin(x)<=1 for all x the sum will converge. The G.P given here has 1st term(a) as sin(x), and common ratio(r) as sin^2(x) so the given sum(S) becomes

S=a/(1−r) =sin(x)/(1−sin^2(x)) = 1……(A)......(which is given to be 1 in the question)

Simplifying.........

sin^2(x)+sin(x)−1=0

This is a quadratic in sin(x) having solutions as: sin(x)=((5^0.5)-1)/2..................(B) (since x is acute so value of sin(x) >0)

From (A), sin(x)/(1-sin^2(x)) =1 ................or ( sin(x) / cos(x) ) =cos(x)..........................

so...........tan(x)=cos(x)...........2tan^8(x)=2(cos^2(x))^4....................(C)

Now from the previous value of sin(x) from (B) the value of cos^2(x)=1-sin^2(x) will be got which should be plugged into (C) to obtain................2tan^8(x)=7-3*(5^0.5)...........................so a=7, b=3, c=5...............a+b+c=15(Answer)

First, simplify the summation given...

(sinx) + (sinx)^3 + (sinx)^5 + ... = (sinx) / (1-(sinx)^2) = sinx / (cosx)^2 = tanx * secx = 1

This also means sinx = (cosx)^2 which we will use later

Now, we need to simplify 2*(tanx)^8. We will use the fact that (tanx)^2 = (secx)^2 - 1 and that tanx * secx = 1

2 (tanx)^8 = 2 (tanx)^4 * ((secx)^4 - 2*(secx)^2 + 1) = 2(tanx)^4(secx)^4 - 4(tanx)^4(secx)^2 + 2(tanx)^4 = 2 - 4(tanx)^2 + 2 - 2(tanx)^2 = 4 - 6(tanx)^2

Almost there. All that's left is to figure out how to evaluate (tanx)^2

From our first summation above, we have sinx = (cosx)^2 = (1 - sinx)^2. If we solve the equation for sinx, we can then solve for (sinx)^2 and (cosx)^2. I'm skipping the algebra but have given (sinx)^2 and (cosx)^2 below.

(sinx)^2 = (6 - 2sqrt5) / 4

(cosx)^2 = 2sqrt5 - 2 / 4

(tanx)^2 = -1/2 + sqrt5 /2

4 - 6 (tanx)^2 = 7 - 3 sqrt5

7+3+5 = 15

$\displaystyle \sum_{n=0}^\infty \sin^{2n+1} \theta =\sin\theta +\sin^3\theta +\cdots =1$ which is a geometric series with common ratio $\sin^2\theta$ . So $\displaystyle \sum_{n=0}^\infty \sin^{2n+1}\theta=\frac {\sin\theta}{1-\sin^2 \theta}=1\Rightarrow \sin\theta=1-\sin^2\theta \Rightarrow$ $\sin^2\theta+\sin\theta-1=0$ . Making the substitution $y=\sin\theta$ , we get $y^2+y-1=0$ . Solving this we get: $y=\frac {\sqrt{5}-1}{2}=\sin\theta$ . Now using $\frac {\sin\theta}{1-\sin^2 \theta}=1$ we get two more facts: $\tan\theta=\cos\theta$ and $\sin\theta=\cos^2\theta$ . Now $2\tan^8\theta=2\cos^8\theta=2\sin^4\theta$ . Finally substituting $\sin\theta=\frac {\sqrt{5}-1}{2}$ into our equation we get $2\tan^8\theta=2\sin^4\theta=7-3\sqrt{5}$ . $7+3+5=15$ .

My mutiplying the sum with $(sinx)^2$ and subtract, noting that $(sinx)^(2n+1)\rightarrow0$ when $n\rightarrow+\infinite$ , we have $sinx/(1-(sinx)^2)=1$ . since x is an acute angle, this gives $sinx=\frac{-1+sqrt5}{2}$ Moreover $sinx=1-(sinx)^2=(cosx)^2$ then $(cosx)^2=\frac{sqrt5-1}{2}$ and \8tanx=cosx)

Hence $2(tanx)^8=2(cosx)^8=2.(\frac{sqrt5-1}{2}^4)=7-3sqrt5$ The answer is 15.

$T_0$ = $\sin\theta$ , $T_1$ = $\sin^3\theta$ , $T_2$ = $\sin^5\theta$ , ............... . . Therefore we conclude that it will make an infinite GP.......... Here We observe that common difference is $sin^2\theta$ ..........We know that SUM of an infinite GP is ^a\(1-r) where r is the common difference in the GP........ Now, $\sin\theta$ divided by $1-(sin^2\theta$ ) is equal to 1(as given in the question){by putting the above statement GP's sum} ..... $\sin\theta$ = $1-(sin^2\theta$ ) {cross-multiplication} ........... Let $\sin\theta$ = y. We make a quadratic equation in y. Equation is : $y^2+y-1$ . From this equation we find the value of $y$ i.e. $\sin\theta$ (by quadratic formula) which is equal to $\frac{-1 \pm \sqrt{5}}{2}$ ............................. We will take the one which has positive sign and neglect the one with negative sign as if we take the one with negative sign, value will be less than $-1$ which cannot be a value of $\sin\theta$ ........ Now, $\sin\theta$ = $\cos^2\theta$ ....... We have to find 2 $\tan^8\theta$ i.e. 2( $\sin^8\theta$ / $\cos^8\theta$ ) which is now 2( $\sin^8\theta$ / $\sin^4\theta$ ) i.e. 2 $\sin^4\theta$ = 2 $y^4$ ................. We find out $y^4$ which is equal to $\frac{7- 3\sqrt{5}}{2}$ {by raising y(which we had got earlier) to power 4} and now multiply it by 2 to get 2 $y^4$ which would come out as $7 - 3\sqrt{5}$ .............. which is in the form of $a - b\sqrt{c}$ .......... therefore the solution is 7+3+5 i.e. $15$ .

We note that $\sum_{i=0}sin^{2n+1}$ ) is the same as sinθ + $sin^{3}$ θ + $sin^{5}$ + ...

Since sinθ is only from (0,1), we can note that this is an infinite geometric progression with common ratio of $sin^{2}$ θ.

So the sum is given by $\frac{sinθ}{1-sin^{2}θ}$ . Then $\frac{sinθ}{1-sin^{2}θ}$ = 1 or $sin^{2}$ θ + sinθ -1 = 0

Solving for sinθ, we get sinθ = $\frac{\sqrt{5}-1}{2}$

We note that sinθ = $cos^{2}$ θ based from the first equation since 1- $sin^{2}$ θ = $cos^{2}$ θ

So $tan^{8}$ = $\frac{sin^{8}θ}{cos^{8}θ}$ = $\frac{sin^{8}θ}{(cos^{2}θ)^{4}}$ = $\frac{sin^{8}θ}{sin^{4}θ}$ = $sin^{4}$ θ

So 2 $tan^{8}$ = 2 $(\frac{\sqrt{5}-1}{2})^{4}$ = 7 - 3 $\sqrt{5}$ ; a + b + c then is 7 + 3 + 5 =15

By summing the geometric series, we get $\begin{aligned}\sin\theta+\sin^3\theta+\sin^5\theta+\cdots&=\frac{\sin\theta}{1-\sin^2\theta}\\&=\frac{\sin\theta}{\cos^2\theta}\\&=1,\end{aligned}$ so $\sin\theta=\cos^2\theta$ .

Let $\tan^2\theta=x$ ; clearly $x\geq0$ . Also, $1+x=\sec^2\theta$ , so $\cos^2\theta=\frac1{1+x}$ and $\sin^2\theta=1-\frac1{1+x}=\frac{x}{1+x}$ . Thus $\begin{aligned}\sin^2\theta&=\cos^4\theta\\\frac{x}{1+x}&=\frac1{(1+x)^2}\\x(1+x)&=1\\x&=\frac{-1\pm\sqrt5}2\end{aligned}$ But $x\geq0$ , so $x=\frac{-1+\sqrt5}2$ . Therefore $2\tan^8\theta=2x^4=7-3\sqrt5$ , and the answer is $7+3+5=\boxed{15}$ .

The sum is \sin\theta+\sin^3\theta+\sin^5\theta....So it is an infinite G.P.and according to the formula the sum is \frac{ \sin\theta}{1-\sin^2\theta} which is given is equal to 1.so we find that as \theta is an acute angle the value of \sin\theta is \frac{\sqrt{5}-1}{2}.hence we can easily find out \tan\theta as well as 2\tan^8\theta.

cosθ=0 does not sastìy the problem

multiply the equation by 2cos θ we have 2cosθ= sin2θ ( 1 + (sinθ)^2 +(sinθ)^4 + .... )

do this one more time we have 4(cosθ)^2=sin2θ ( 2cosθ + sin2θ*( sinθ + (sinθ)^3 +......))

<=> 4(cosθ)^2= sin2θ* (2cosθ +sin2θ)

<=>4(cosθ)^2 = 4sinθ * (cosθ)^2 + 4 (sinθcosθ)^2

<=> (sinθ)^2 + sinθ -1=0 (*)

<=> sinθ= (cosθ)^2 <=> (sinθ)^4 = (cosθ)^8

(tanθ)^8= (sinθ)^8/ (cosθ)^8= (sinθ)^4

(*)<=> sin θ = ( - 1 + V5 )/2 => (sinθ)^2= (3 - V5)/2 => (sinθ)^4= (7 - 3V5)/2 =>2(tanθ)^8=7 -3V5

a+b+c = 7+3+5 =15

$\text{Let }s=\sin \theta \text{ and }t=\tan\theta$

$\frac{s}{1-s^2}=1\Rightarrow s=\frac{-1+\sqrt{5}}{2}$

$s^2=\frac{3-\sqrt{5}}{2}$

$t^2=\frac{(3-\sqrt{5})/2}{1-(3-\sqrt{5})/2}=\frac{\sqrt{5}-1}{2}$

$2t^2=\sqrt{5}-1$

$4t^4=6-2\sqrt{5}$

$16t^8=56-24\sqrt{5}$

$2t^8=7-3\sqrt{5}$

∑(n=0 to ∞)sin^(2n+1)θ=sinθ+ sin³θ+ sin⁵θ+ sin⁷θ+ ....... ∞ we observe it si infinite G.P series with common Ratio sin²θ it its value always lies (0 1) so |r|<1 then sum =sinθ/(1-sin²2θ)=1 OR cos²θ= sinθ...................(1)

OR sin²θ+sinθ-1=0-----------------(2) ON SOLVING WE GET sinθ= (√5 - 1)/2 (NEGATIVE PART REJECTED) NOW 2tan⁸θ= 2 sin⁸θ/cos⁸θ = 2 sin⁸θ/(cos²θ)⁴ =2 SIN^4θ FROM (1) =2 [ (√5 - 1)/2]^4= (7 - 3√5 ) SO ANSWER IS 7+3+5=15

As sequence will be similar to an infinite GP , so $\frac { \sin \theta}{1 - \sin^2\theta}$ =1 On solving the quadratic we get values of $sin\theta$ and hence $tan\theta$ = $\sqrt{(\sqrt{5} - 1)/2}$

ANd now on simplifying $2\tan^8\theta$ we get $7 -3\sqrt{5}$

It's easy to see that $\sum_{n = 0}^{\infty}\sin^{2n + 1}\theta$ is a infinity geometric progression of ratio $\sin^2\theta$ , which is clearly lesser than 1 so, we have: $$ S = \frac{a 1}{1 - q} \Rightarrow \sum {n = 0}^{\infty}\sin^{2n + 1}\theta = \frac{\sin\theta}{1 - \sin^2\theta} = 1 $$ Using some trigonometric transformations we'll get to: $$ \tan\theta\sec\theta = 1 \Rightarrow \tan\theta\sqrt{1 + \tan^2\theta} = 1 $$ Squaring both sides: $$ \tan^2\theta(1 + \tan^2\theta) = 1 \Rightarrow \tan^4\theta + \tan^2\theta - 1 = 0 $$ Solving the quadratic equation for $\tan^2\theta$ we'll get to: $$ \tan^2\theta = \frac{-1 \pm \sqrt{5}}{2} $$ Which only $\frac{-1 + \sqrt{5}}{2}$ is positive, squaring $\tan^2\theta$ twice we'll get close to the solution: $$ \tan^8\theta = \frac{7 - 3\sqrt{5}}{2} $$ Passing the $2$ to the left side we finally get to the required solution: $$ 2\tan^8\theta = 7 - 3\sqrt{5} $$ Finally: $$ a + b + c = 7 + 3 + 5 = 15 $$

The given equation can be reduced using the infinite geometric series formula to $\displaystyle \frac{\sin\theta}{\cos^2\theta}=1$ .

Using basic trigonometric identities, this is equivalent to $\displaystyle \tan^2\theta-\frac{1}{\tan^2\theta}.$

Squaring, we have $\displaystyle \tan^4\theta+\frac{1}{\tan^4\theta}=3.$

Squaring again, $\displaystyle \tan^8\theta+\frac{1}{\tan^8\theta}=7.$

This last equation is can be converted to a quadratic in $\tan^8\theta$ , and solving gives $2\tan^8\theta=7-3\sqrt{5}$ , so the answer is 7+3+5=15.

Since $\theta$ is an acute angle we know that $\sin \theta \in (0,1)$ and so the geometric progression $\sum_{n=0}^\infty \sin^{2n+1}\theta$ converges and is equal to $\frac{\sin \theta}{1 - \sin^2\theta} = \frac{\sin\theta}{\cos^2\theta}$ . We've been told that this value is equal to $1$ so we have $1 - \sin^2 \theta = \sin \theta$ and solving this as a quadratic for $\sin \theta$ gives $\sin \theta = \frac{-1 \pm \sqrt{5}}{2}$ . Since $\theta$ is acute it must be the positive root so $\sin \theta = (\sqrt 5 - 1)/2$ . Now we know that $\tan^2\theta = \frac{\sin^2 \theta}{\cos^2\theta} = \sin\theta \cdot \frac{\sin \theta}{\cos^2 \theta} = \sin \theta$ . So $2 \tan^8 \theta = 2 sin^4 \theta = 2 (\sqrt 5 - 1)^4/16 = 7 - 3 \sqrt{5}$ . Thus the answer is $3 + 5 + 7 = \boxed{15}$ .

Given series is as , $\sin \theta + \sin ^{3} \theta +\sin ^{5} \theta + ...$

it is a GP with a= $\sin \theta$ & r = $\sin ^{2} \theta$ so $\frac{a}{1-r} = \frac{\sin \theta }{1-\sin ^{2} \theta} = 1$ $\sin ^{2} \theta + \sin \theta - 1 = 0$ ... (1)

Also, $\sin \theta = \cos ^{2} \theta$ .... (2)

on putting x = $\sin \theta$ , we get $x^{2} + x + 1 = 0$

and we get x = $\frac{-1 +- \sqrt{5}}{2}$

from which only x = $\frac{-1 + \sqrt{5}}{2}$ is acceptable

Now 2 $\tan ^{8}\theta = (\frac{\sin ^{2} \theta}{\cos ^{2} \theta}) ^ 4$

on using ...(2) we get

2 $\tan ^{8}\theta = (\sin {2} \theta) ^ {4}$

= $x^{4}$

= $7 - 3\sqrt{5}$

Hence a + b +c = 7 + 3 + 5 = 15

∑(n=0 to ∞)sin^(2n+1)θ=sinθ+ sin³θ+ sin⁵θ+ sin⁷θ+ ....... ∞

we observe it Si infinite G.P series with common Ratio sin²θ it its value always lies (0 1) so |r|<1

then sum =sinθ/(1-sin²2θ)=1 OR cos²θ= sinθ...................(1)

OR sin²θ+sinθ-1=0-----------------(2)

ON SOLVING WE GET sinθ= (√5 - 1)/2 (NEGATIVE PART REJECTED)

NOW 2 tan⁸θ= 2*sin⁸θ/cos⁸θ

= 2*sin⁸θ/(cos²θ)⁴

=2*SIN^4θ FROM (1)

=2*[ (√5 - 1)/2]^4=

(7 - 3√5 )

SO ANSWER IS 7+3+5=15

The solution what I have arrived at is: a + b + c = 7 + 3 + 5 = 15

i) Evaluating the given summation ∑sin²ⁿ+¹(Ө) for n = 0 to ∞, This sum S = sin(Ө) + sin³Ө + sin⁵Ө + sin⁷Ө + ....... up to infinity

This is in GP, whose 1st term is sin(Ө) and common ratio is sin²Ө As Ө is an acute angle, sin(Ө) is in (0, 1); so, sin²Ө also in (0, 1) ==> |r| < 1 So the series sum forms infinite sum, which equal = a/(1 - r) = sinӨ/(1 - sin²Ө)

ii) Given sum = 1 ==> sinӨ = 1 - sin²Ө = cos²Ө ------ (1) Rearranging, sin²Ө + sinӨ - 1 = 0

This is a quadratic in sinӨ; solving sin(Ө) = (√5 - 1)/2 ----- (2) [Only positive values is taken; negative value is rejected, since Ө is acute]

iii) tan⁸Ө = sin⁸Ө/cos⁸Ө = sin⁸Ө/(cos²Ө)⁴ But from (1), cos²Ө = sinӨ

Substituting this in the above, it reduces to: tan⁸Ө = sin⁸Ө/sin⁴Ө = sin⁴Ө ----- (3)

iv) From (2), sin(Ө) = (√5 - 1)/2 ==> sin²Ө = [(√5 - 1)/2]² = (3 - √5 )/2

==> sin⁴Ө = [(3 - √5 )/2]² = (7 - 3√5 )/2

v) Hence of the above, 2 tan⁸Ө = 2 sin⁴Ө = 2*[ (7 - 3√5 )/2] = (7 - 3√5 )

==> a−b√c = 7 - 3√5; so a = 7; b = 3 and c = 5

Hence a + b + c = 15

Write out the series:

$\sin \theta$ + $\sin^{3} \theta$ + $\sin^{5} \theta$ + $\sin^{7} \theta +... = 1$

Factor out the sine.

$\sin \theta$ $(1$ + $\sin^{2} \theta$ + $\sin^{4} \theta$ + $\sin^{6} \theta +...) = 1$

Let $u=\sin \theta$ .

$u (1+u^{2}+u^{4}+u^{6}+...) = 1$

It's a geometric series now, with common ratio $u^2$ .

$u (\frac{1}{1-u^{2}}) = 1$

$u^{2}+u-1=0$

Solve with quadratic formula to get:

$\sin \theta = \frac{-1+\sqrt{5}}{2}$

Only use the positive value because sine ranges from -1 to 1.

Use identities:

$\sin^{2} \theta + \cos^{2} \theta = 1$

$(\frac{-1+\sqrt{5}}{2})^{2} + \cos^{2} \theta = 1$

$\cos^{2} \theta = \frac{-1+\sqrt{5}}{2}$

$\tan^{2} \theta = \frac{1}{\cos^{2} \theta} -1$

$\tan^{2} \theta = \frac{2}{-1+\sqrt{5}} -1$

Square it twice to get

$\tan^{8} \theta = \frac{7-3 \sqrt{5}}{2}$

$2 \tan^{8} \theta = 7 - 3 \sqrt{5}$

$7+3+5=15$

There is no need to expand the expression to find the values. We can see that $\frac{\sin \theta}{1-\sin^2 \theta}=1$ .

From this we get $\tan \theta=\cos \theta$ .

So 2 $(\tan \theta)^8=2(\cos \theta)^8$

=2 $(1-\sin^2 \theta)^4$ =2 $\sin^4 \theta$ = $2(1-\sin \theta)^2$ . From this we can get the final expression easily.

We notice that the sum is a geometric series $\sin\theta + \sin^3\theta + \sin^5\theta + \dots = 1$ The sum of this infinite series can be expressed as $\dfrac{\sin\theta}{1 - \sin^2\theta} = 1$ which can be rearranged to the quadratic $\sin^2\theta + \sin\theta - 1 = 0$ the positive solution is $\sin\theta = \dfrac{\sqrt{5} - 1}{2}$ Notice also that $1 - \sin^2\theta = cos^2\theta$ therefore $sin\theta = cos^2\theta$ and $2\tan^8\theta\ = 2\dfrac{\sin^8\theta}{(\cos^2\theta)^4} = 2\dfrac{\sin^8\theta}{\sin^4\theta} = 2\sin^4\theta = 2(\dfrac{\sqrt{5} - 1}{2})^4 = 7 - 3\sqrt{5}$ a + b + c = 15

The given equation is:

     sin x+sin^3 x+sin^5 x....=1

     sin x + sin^2 x ( sin x + sin^3 x + sin^5 )=1

     which gives,sin x+sin^2 x=1  or  sin x=cos^2 x   or  tan x = cos x

     on solving sin x = (√5−1)/2 = tan^2 x

     on squaring two times, we get tan^8 x= (7−3√5)/2

So, 2 tan^2 x=7−3√5

So a+b+c=15

We have: $\sin\theta + \sin^3\theta+ \sin^5\theta+...=1$ .

Therefore, $\sin\theta+\sin^2\theta( \sin\theta + \sin^3\theta+...)=1$ or $\sin\theta +\sin^2\theta=1$ .

So $\sin\theta=\frac{\sqrt{5}-1}{2}$ .

Because $\tan\theta=\frac{\sin\theta}{\sqrt{1-\sin^2\theta}}$ $=\frac{\sin\theta}{\sqrt{\sin\theta }}=\sqrt{\sin\theta}$

Hence $\tan^8\theta=\sin^4\theta=7-3\sqrt{5}$ .

So the answer is: 7+3+5=15