Question

Please help, performance task: trigonometric identities

Answer 1

Answer:

Trigonometric Identities used:

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

$\sec \theta \equiv \dfrac{1}{\cos \theta}$

$\cot \theta \equiv \dfrac{1}{\tan \theta}$

$\sec^2 \theta \equiv 1 + \tan^2 \theta$

$\sin(\theta)=\cos \left(\dfrac{\pi}{2}-\theta\right)$

Part (a)

   $(1+ \cos(x))(1-\cos(x))$

$=1-\cos(x)+\cos(x)-\cos^2(x)$

$=1-\cos^2(x)$

$= \sin^2(x)$

Part (b)

   $\dfrac{1}{\cot^2(x)}-\dfrac{1}{\cos^2(x)}$

$=\tan^2(x)-\sec^2(x)$

$=\tan^2(x)-(1 + \tan^2(x))$

$=\tan^2(x)-1 - \tan^2(x)$

$=-1$

Part (c)

   $\sec^2\left(\dfrac{ \pi }{2}-x\right) \left[\sin^2(x)-\sin^4(x) \right]$

$=\dfrac{1}{\cos^2\left(\dfrac{ \pi }{2}-x\right)} \left[\sin^2(x)-\sin^4(x) \right]$

$=\dfrac{1}{\sin^2(x)} \left[\sin^2(x)-\sin^4(x) \right]$

$= \dfrac{\sin^2(x)-\sin^4(x)}{\sin^2(x)}$

$= \dfrac{\sin^2(x)}{\sin^2(x)} - \dfrac{\sin^4(x)}{\sin^2(x)}$

$= \dfrac{\sin^2(x)}{\sin^2(x)} - \dfrac{(\sin^2(x))(\sin^2(x))}{\sin^2(x)}$

$=1- \sin^2(x)$

$= \cos^2(x)$

Answer 2

#a

• (1+cosx)(1-cosx)
• 1²-cos²x
• 1-cos²x
• sin²x

#2

• 1/cot²x-1/cos²x
• tan²x-sec²x
• -1

#c

• sec²(90-x)[sin²x-sin⁴x]
• csc²x[sin²x-sin⁴x]
• 1-sin²x
• cos²x