Squared Trigonometric Functions

If $\sin \theta = \frac{-12}{13}$ then using the identity, $\cos\theta = \frac{-5}{13}$ . $\cos \theta$ is negative because $\theta$ is in the third quadrant.

$180(\sin^{2} \frac{\theta}{2} + \cos^{2} \frac{\theta}{2} + \tan^{2} \frac{\theta}{2})$ $=180(1 + \tan^{2} \frac{\theta}{2}) =18( \sec^{2} \frac{\theta}{2})$ $= 180(\frac{1}{\cos^{2} \frac{\theta}{2}})$ $= 180(\frac{2}{1 + \cos \theta})$

Now substitute the value and you get 585.

Solving for the value of $\cos^2\frac{\theta}{2}$ ,

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

$\Rightarrow(\frac{-12}{13})^2+\cos^2\theta=1$

$\Rightarrow\cos^2\theta=\frac{25}{169}$

$\Rightarrow\cos\theta=\frac{-5}{13}$ , since $180^{\circ} < \theta < 270^{\circ}$ .

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

$\cos^2\frac{\theta}{2}=\dfrac{1+(\dfrac{-5}{13})}{2}$

$\cos^2\frac{\theta}{2}=\frac{4}{13}$

We can simplify the expression in the parenthesis:

$180(1+\tan^2\frac{\theta}{2})$ , since $\sin^2\frac{\theta}{2}+\cos^2\frac{\theta}{2}=1$

$\Rightarrow 180(\sec^2\frac{\theta}{2})$

$\Rightarrow 180(\dfrac{1}{\cos^2\dfrac{\theta}{2}})$

$\Rightarrow 180(\frac{13}{4})$

$\Rightarrow \boxed{585}$ .

>We know that:

> $\boxed{ \sin^{2} \beta + \cos^{2} \beta = 1}$

> $\sin^{2} \frac{\theta}{2} + \cos^{2} \frac{\theta}{2} = 1$ , thus $180 \times (1+ \tan^{2} \frac{\theta}{2})$

> If $\sin \theta = \frac{-12}{13}$ then $\cos \theta = ^{+}_{-}\frac{5}{13}$ but $\theta$ belongs to the 3rd quadrant, so $\cos \theta = \frac{-5}{13}$ ,

>and $\tan \theta = \frac{12}{5}$

>thinking in sum of arcs, we can see:

> $\boxed{ \tan 2\times \beta = \frac{2 \times \tan \beta}{1 - \tan^{2} \beta} }$ applying in this case, $\tan 2\times \frac{\theta}{2} = \frac{2 \times \tan \frac{\theta}{2}}{1 - \tan^{2} \frac{\theta}{2}}$

> replacing $\tan 2\times \frac{\theta}{2}$ for $\tan \theta$ and $\tan \frac{\theta}{2}$ for "x", we got: $\frac{12}{5}= \frac{2 \times x}{1 - x^{2}}$

> solving, we have: $6x^{2} + 5x - 6 = 0$ with $x^{'}=\frac{-18}{12}$ and $x^{''}=\frac{8}{12}$ , but $\frac{\theta}{2}$ belongs to the 2nd quadrant, so $\tan \frac{\theta}{2} < 0$ thus $\boxed{ \tan \frac{\theta}{2} = \frac{-18}{12} }$

> $180 \times (1+ \tan^{2} \frac{\theta}{2})$ = $180 \times (1+ (\frac{-18}{12})^2)$ = $\boxed{585}$

sin²ϕ + cos²ϕ = 1,
sinϴ=-12/13 (given) => thus cosϴ = -5/13 (third quadrant)

tanϴ/2 = sinϴ/(1+cosϴ) => half angle identity.
Therefore tanϴ/2 ={ (-12/13) ÷( 13/13 + -5/13)} = (-12/13) ÷ (8/13) = -12/8

tan²ϴ = ( -12/8) = 144/64

180(sin²ϴ/2 + cos²ϴ/2 + tan² ϴ/2) = 180(1 + tan² ϴ/2) = 180(1 + 144/64)

180(64/64 + 144/64) = 180(208/64) = 180(3.25) = 585

A.Quinones Cidra, Puerto Rico

the angle is ( a )in 3th quadrant so, cos a has negative value cos a= sqrt(1-sin^2 a)= -5/13

use sin^2 a/2=(1-cosa)/2

cos^2 a/2= (1+cos a)/2 and tan^2 a/2= (1-cos a)/(1+cos a)

you will find 585 ^^

If (theta) is in third quadrant, then cos (theta) is negative. Cos(theta) = -5/13.

Using trigonometric identities, sin (theta/2) = 3sqrt(13)/13

cos (theta/2) = 2sqrt(13)/13

tan (theta/2) = -3/2

Squaring them gives 9/13, 4/13, and 9/4 respectively. Adding them and multiplying to 18 gives 585.

Sin^{2} \frac{/theta}{2} + Cos^{2} \frac{/theta}{2} =1 Tan^{2} \frac{/theta}{2} = /frac{1-Cos/theta} {1+Cos/theta}

if sin θ is -ve value it's mean that it's in third or forth quarter but we have a condition that said 180 < θ < 270 ... that's mean θ is in third one when we get the angle sin^{-1} ( -12/13) it will be = \boxed{-67.38}

the -ve mean third quarter and that mean the angle = 67.38+180 = \boxed{247.38}

247.38/2 = \boxed{123.69}

our equation will be ... 180[sin^{2} 123.69 + cos^{2} 123.69 + tan^{2} 123.69 ] = 585

The value of THITA is Sine inverse (-12/13). But it will have to be within 180<thita<270. So it will be approximately 247.3801351. Then put into equation. Get the answer 585 :)

sinθ=-12/13 then cos θ=-5/13

180(sin^2θ/2+cos^2θ/2+tan^2θ/2)=

180(1+tan^2θ/2)=180sec^θ/2

180(1/(1+cosθ)/2)=360/(1+cosθ)=

360/(1-5/13)=585