Ryan's Sines and Cosines

We are Given $sin2 \theta$ = $\frac {2}{3}$

$\Rightarrow$ 2 $sin \theta cos \theta$ = $\frac {2}{3}$

$\Rightarrow$ $sin \theta cos \theta$ = $\frac {1}{3}$

Now we may write $sin^6 \theta$ as $(sin^2 \theta)^3$ and $cos^6 \theta$ as $(cos^2 \theta)^3$ .

Now applying Identity $A^3 + B^3 = (A+B)^3 - 3AB(A+B)$ , we have:

$(sin^2 \theta)^3$ + $(cos^2 \theta)^3$ = $(sin^2 \theta + cos^2 \theta)^3 - 3sin^2 \theta cos^2 \theta (sin^2 \theta + cos^2 \theta)$

But we know that $(sin^2 \theta + cos^2 \theta) = 1$

So the above expression reduces to:

$(1)^3-3(sin^2 \theta cos^2 \theta)(1)$

But we deduced above that $sin \theta cos \theta$ = $\frac {1}{3}$

$\Rightarrow$ $sin^2 \theta cos^2 \theta$ = $\frac {1}{9}$

So on substituting and solving we get $sin^6 \theta + cos^6\theta = \frac {2}{3}$

So $a+b = 2+3 = 5$

Let n = sin^6 (θ) + cos^6 (θ). After looking at it, you will realize that it is a sum of two cubes.

n = [sin^2 (θ)]^3 + [cos^2 (θ)]^3

Thus, it can be factored into this:

n = [sin^2 (θ) + cos^2 (θ)]*[sin^4 (θ) +cos^4 (θ) -cos^2 (θ) sin^2 (θ)]

n = [1]*[sin^4 (θ) +cos^4 (θ) -cos^2 (θ) sin^2 (θ)]

n = sin^4 (θ) +cos^4 (θ) -cos^2 (θ) sin^2 (θ)

After completing the square we get,

n = [sin^2 (θ) + cos^2 (θ)]^2 -3cos^2 (θ) sin^2 (θ)

n = 1^2 - 3cos^2 (θ) sin^2 (θ)

n = 1 - 3cos^2 (θ) sin^2 (θ)

(Equation 1)

From the given, sin 2θ = 2sinθcosθ = 2/3

Thus, sinθcosθ = 1/3.

Squaring both sides and multiplying each by 3 we get,

3sin^2 (θ) cos^2 (θ) = 1/3.

(Equation 2)

Substituting Equation 2 into Equation 1,

n = 1 - 1/3

n = 2/3

Thus, a+b = 2+3 = 5

We have, \sin \2theta = \frac {2}{3}

or, 2 \sin \theta \cos \theta = \frac {2}{3}

or, \sin \theta \cos \theta = \frac {1}{3} ..... (i)

Now, (\sin \theta)^6 + (\cos \theta)^6

= [(\sin \theta)^2]^3 + [(\cos \theta)^2]^3

= [(\sin \theta)^2 + (\cos \theta)^2]^3 - 3 \times [(\sin \theta)^2(\cos \theta)^2] \times [(\sin \theta)^2 + (\cos \theta)^2]

=1^3 - 3 \times (\sin \theta \cos \theta)^2 \times 1

=1- 3 \times \frac {1}{3} \times 1

=1 - \frac {1}{3}

= \frac {2}{3}

So, our answer is 2+3 = 5.

Given $\sin \theta = \frac {2}{3}$

We get $\sin \theta \cos \theta= \frac {1}{3}$

Then $\sin^6 \theta + \cos^6 \theta$ $= (\sin^3 \theta + \cos^3 \theta )^2 - 2 \sin^3 \theta \cos^3 \theta$ $= [(\sin^2 \theta + \cos^2 \theta )( \sin \theta + \cos \theta ) - (\sin \theta \cos \theta )( \sin \theta + \cos \theta )]^2 - 2 \sin^3 \theta \cos^3 \theta$ $= [ \frac {2}{3} (\sin \theta + \cos \theta)]^2 - 2(\sin \theta \cos \theta)^3$ $= \frac {4}{9} (1 + \frac {2}{3}) - \frac {2}{27}$ $= \frac {4}{9} + \frac {8}{27} - \frac {2}{27}$ $= \frac {18}{27} = \frac {2}{3} = \frac {a}{b}$

Hence, $a + b = 2 + 3 =5$

As we can see: $$ (a^3 + b^3) = (a + b)(a^2 - ab + b^2) $$ and: $$ (a^2 + b^2) = (a + b)^2 - 2ab $$ Thus: $$ (a^3 + b^3) = (a + b)( (a+ b)^2 - 3ab) $$ using a and b as: $$ a = \sin^2 \theta \ and \ b = \cos^2\theta $$ then $$ (\sin^6\theta + \cos^6\theta) = (\sin^2\theta + \cos^2\theta)((\sin^2\theta + \cos^2\theta) - 3\cdot\sin^2\theta\cos^2\theta) $$ Now, knowing that $$ \sin^2\theta + \cos^2\theta = 1 $$ and $$ 2\sin\theta\cos\theta = \sin(2\theta) $$ Hence: $$ (\sin^6\theta + \cos^6\theta) = 1\cdot(1 - \frac{3}{4}\sin^2(2\theta)) $$ Since: $$ \sin^2(2\theta) = \frac{2}{3} $$ We finally came to a conclusion: $$ (\sin^6\theta + \cos^6\theta) = 1 - \frac{3}{4}\cdot\frac{4}{9} = 1 - \frac{1}{3} = \frac{2}{3} = \frac{a}{b} $$ Solution: $$ a + b = 5 $$

From $\sin(2\theta) = \frac {2} {3}$ , we deduce - $\sin(2\theta) = 2 \sin \theta \cos \theta = \frac {2} {3}$ $\Rightarrow \sin \theta \cos \theta = \frac {1} {3}$

Now solving for $\sin^6 \theta + \cos^6 \theta$ , we get - $\sin^6 \theta + \cos^6 \theta = (\sin^2 \theta)^3 + (\cos^2 \theta)^3$

$\Rightarrow(\sin^2 \theta + \cos^2 \theta)\cdot(\sin^4 \theta + \cos^4 \theta - \sin^2 \theta\cos^2 \theta)$

By using $\sin^2 \theta + \cos^2 \theta = 1$ , we get -

$\Rightarrow(\sin^4 \theta + \cos^4 \theta - \sin^2 \theta\cos^2 \theta)$

$\Rightarrow((\sin^4 \theta + \cos^4 \theta + 2\sin^2 \theta\cos^2 \theta) - 3\sin^2 \theta\cos^2 \theta)$

$\Rightarrow (\sin^2 \theta + \cos^2 \theta)^2 - 3\sin^2 \theta \cos^2 \theta$

Again by using $\sin^2 \theta + \cos^2 \theta = 1$ we get -

$\Rightarrow 1 - 3\sin^2 \theta \cos^2 \theta$

But we have already calculated the value of $\sin \theta\cos \theta$ i.e $\frac {1} {3}$ , so -

$\Rightarrow 1 - 3(\frac {1} {3})^2$

$\Rightarrow 1 - \frac {1} {3}$

$\Rightarrow \frac {2} {3}$

Therefore , $\sin^6 \theta + \cos^6 \theta = \frac {a} {b} = \frac {2} {3}$

Hence , $a + b = 5$

sin 2theta = 2/3 = 2sin theta cos theta,or sin theta cos theta = 1/3 ( sin^2 theta + cos^2 theta)^3=1 sin^6 theta + cos^6 theta + 3sin^2theta cos^2theta(sin^2 theta + cos^2 theta)=1 sin^6 theta + cos^6 theta + 3(sin theta cos theta)^2(1)=1 sin^6 theta + cos^6 theta + 3 (1/3)^2=1 sin^6 theta + cos^6 theta +3 1/9=1 sin^6 theta + cos^6 theta +1/3=1 sin^6 theta + cos^6 theta =1-1/3 sin^6 theta + cos^6 theta =2/3 therefore 2+3 =5(answer)

we know that, \sin\theta^6 + \cos\theta^6 = (\sin\theta^2)^3 + (\cos\theta^2)^3 = (\sin\theta^2 + \cos\theta^2)(\sin\theta^4 + \cos\theta^4 - \sin\theta^2\cos\theta^2) = 1(\sin\theta^2 + \cos\theta^2)^2 - 2\sin\theta^2\cos\theta^2) - \sin\theta^2\cos\theta^2) = (1 - 3\sin\theta^2\cos\theta^2) = (1 - 4\sin\theta^2\cos\theta^2 + \sin\theta^2\cos\theta^2 Multiplying and dividing throughout by 4, = (4 - 16\sin\theta^2\cos\theta^2 + 4\sin\theta^2\cos\theta^2)\times1/4 = {4 - 4[(\sin\2theta)^2] + (\sin\2theta)^2}\times1/4 = (4 - 4(4/9) + 4/9)\times1/4 = (4 - 16/9 + 4/9)\times1/4 = (24/9)\times1/4 = 2/3 Ans: 2/3

sin^{6}θ+cos^{6}θ=\frac {a}{b} We know, sin2θ=2sinθ*cosθ and sin^{2}θ+cos^{2}θ=1

(1-cos^{2}θ)^{3}+cos^{6}θ=\frac {a}{b} [(a-b)^{3}=a^3-b^3-3ab(a-b)] or (1-cos^{6}θ-(3 cos^{2}θ (1-cos^{2}θ)) +cos^{6}θ=\frac {a}{b} or 1-(3 cos^{2}θ (1-cos^{2}θ))=\frac {a}{b} or 1-(3 cos^{2}θ sin^{2}θ)=\frac {a}{b} or 1-(.75 (2cosθ sinθ)^{2})=\frac {a}{b} or 1-(.75 (sin2θ)^{2})=\frac {a}{b} or 1-(3/4 2/3*2/3)=\frac {a}{b} or 1-(1/3)=\frac {a}{b} or 2/3=\frac {a}{b}

Therefore, a=2 and b=3 so, a+b=5

$sin(2\theta) = \frac{2}{3}$

Using the double angle formula, which states that $sin(2\theta) = 2sin(\theta)cos(\theta)$ , we get $2sin(\theta)cos(\theta) = \frac{2}{3}$ ------> $sin(\theta)cos(\theta) = \frac{1}{3}$

Squaring this, we get $sin^2(\theta)cos^2(\theta) = \frac{1}{9} \dots \dots \dots \dots [1]$

The pythagorean identity states that $sin^2(\theta) + cos^2(\theta) = 1$ . Squaring both sides, we get $sin^4(\theta) + cos^4(\theta) + 2sin^2(\theta)cos^2(\theta) = 1$

Substitute $2 \times [1]$ into $2sin^2(\theta)cos^2(\theta)$ , so $sin^4(\theta) + cos^4(\theta) = \frac{7}{9}$

Multiply the equations $sin^2(\theta) + cos^2(\theta) = 1$ and $sin^4(\theta) + cos^4(\theta) = \frac{7}{9}$ to get

$sin^6(\theta) + cos^6(\theta) + sin^4(\theta)cos^2(\theta) + cos^4(\theta)sin^2(\theta) = \frac{7}{9}$

Note that $sin^4(\theta)cos^2(\theta) + cos^4(\theta)sin^2(\theta)$ can be factorised to $(sin^2(\theta)cos^2(\theta)) \times (sin^2(\theta) + cos^2(\theta))$

$sin^2(\theta)cos^2(\theta) = \frac{1}{9}$ as of [1] and $sin^2(\theta) + cos^2(\theta) = 1$

Therefore, $sin^6(\theta) + cos^6(\theta) + \frac{1}{9} = \frac{7}{9} = \frac{6}{9} = \frac{2}{3}$

Then the answer is $2 + 3 = \boxed{5}$

I hope you guys find it helpful. If you have lemons...