Question

10. Rewrite with only sin x and cos x. (1 point)
sin 2x - cos 2x


a) 2 sinx cosx - 1 + 2 sin2x
b) 2 sin x cos2x - 1 + 2 sin2x
c) 2 sin x cos2x - sin x + 1 - 2 sin2x
d) 2 sin x cos2x - 1 - 2 sin2x

11. Find the exact value by using a half-angle identity. (1 point)
sin 22.5°


a) negative one half times the square root of quantity two plus square root of two.
b) one half times the square root of quantity two plus square root of two.
c) negative one half times the square root of quantity two minus square root of two.
d) one half times the square root of quantity two minus square root of two

15. Verify the identity. (1 point)
cot x minus pi divided by two. = -tan x

NEED HELP ASAP PLEASEEEE!!!

Answer 1

10. Rewrite with only sin x and cos x.
sin 2x - cos 2x

sin2x = 2sinxcosx
cos2x = (cosx)^2 - (sinx)^2 = 2(cosx)^2 -1 = 1- 2(sinx)^2

sin2x- cos2x=2sinxcosx-(1- 2(sinx)^2=2sinxcosx-1+2(sinx)^2

sin2x- cos2x=2sinxcosx-1+2(sinx)^2

the answer is the letter b) 2 sin x cos2x - 1 + 2 sin2x

11. Find the exact value by using a half-angle identity. sin 22.5°Using the half angle formula you get:sin2(θ)=12[1−cos(2θ)]if θ=22.5° then 2θ=45°so you get:sin2(22.5°)=12[1−cos(45°)]sin2(22.5°)=12[1−√2/2]=2−√24and square root both sides:sin(22.5°)=±√2−√24=±0.382so 
sin(22.5°)=0.382the answer is the letter D) one half times the square root of quantity two minus square root of two
15. Verify the identity. 
cot x minus pi divided by two. = -tan x
Cot(x-pi/2)=-tan(x)
 sin(A − B) = sin A cos B − cos A sin B
sin(x – pi/2) = sin x cos (pi/2) − cos x sin (pi/2)=-cosx
 
cos(A − B) = cos A cos B − sin A sin B
cos(x− pi/2) = cos x cos pi/2 − sin x sin pi/2=-sinx
 Cot(x-pi/2)=cos(x-pi/2)/sin(x-pi/2)
= (-sinx)/(-cosx)=-tanx--------------ok

Answer 2

Trigonometric functions can only be applied to right triangles. There are six trigonometric functions. The answer to part 1 is an option (a) and the answer to part 2 is an option (d).

Part (1)

Rewrite with only sin x and cos x.

The given trigonometric expression is $\sin 2x - \cos 2x$.

Apply the following identities in the given expression and solve them further.

$\begin{aligned}\sin2x&=2\;\rm{sin\;x\;cos\;x}\\\cos2x&=1-2\sin^2x\end{aligned}$

Thus,

$\begin{aligned} \sin 2x - \cos 2x&=2 \sin x \cos x-\left(1-2 \sin^2 x \right)\\&=2 \sin x \cos x-1+2 \sin^2 x\\&=2 \sin^2 x+2 \sin x \cos x-1\end{aligned}$

Hence, option (a) is correct.

Part (2)

Find the exact value by using a half-angle identity.

$\sin 22.5$

Now, we know that,

$\begin{aligned}\cos 2x&=1-2 \sin^2 x\\2 \sin^2 x&=1 -\cos 2x\\\sin^2 x&=\dfrac{1-\cos 2x}{2}\\\sin x&= \sqrt {\dfrac{1-\cos 2x}{2}} \end{aligned}$

From the above question, substitute the value of x=22.5, and solve it further.

$\begin{aligned}\sin 22.5&= \sqrt {\dfrac{1-\cos 2 \times 22.5}{2}}\\&= \sqrt {\dfrac{1-\cos 45}{2}}\\&=\sqrt {\dfrac{1- \dfrac{1}{\sqrt 2}}{2}}\\&=\sqrt{\dfrac{\sqrt2-1}{2 \sqrt 2}}\end{aligned}$

Part (3)

Incomplete question.

To know more about it, please refer to the link:

https://brainly.com/question/13041113