If cos(theta) = -sin(theta), find the value of sin^2(theta)

[D!ddy]

Hi, this question here is a multiple choice question. If Cos? = -sin?, then sin[sup:1w3v6pvw]2[/sup:1w3v6pvw]? is?

A. 1/4
B. 1/2
C. sqrt2/2
D. -sqrt2/2

I don't want the answer I just want to know how to go about solving this question, cause I just can't wrap my head around it, I even tried to use the answers to somehow solve the question, but I just don't know what to do with the numbers :lol:

 

[arthur ohlsten]

Re: Trig Question

cos@=-sin@

I would try to determine a value of @. To do that I would try to determne something like Sin@= a value
or cos @ = a value OR
tan@= a value

If I knew @ then I would find sin^2@ it probably = 1/2

ARTHUR
IF YOU STILL HAVE TROUBLE RESUBMIT THE PROBLEM.

 

[Loren]

Re: Trig Question

If the problems said cosA = sinA, would you determine that A was in quadrant I or quadrant III? Draw a sketch and see if you can determine the measure of A when it is in QI. Then do some thinking. I think you will be able to determine the correct answer.

 

[Deleted member 4993]

Re: Trig Question

Hi, this question here is a multiple choice question. If Cos? = -sin?, then sin[sup:260oezvs]2[/sup:260oezvs]? is?

A. 1/4
B. 1/2
C. sqrt2/2
D. -sqrt2/2

I don't want the answer I just want to know how to go about solving this question, cause I just can't wrap my head around it, I even tried to use the answers to somehow solve the question, but I just don't know what to do with the numbers :lol:

Square both sides and get an equation..............................(1)

Then use the fact that:

\(\displaystyle \cos^2(\theta) \, + \, \sin^2(\theta) \, = \, 1\) ...........(2)

combining (1) and (2) you should get your answer.

 

[soroban]

Re: Trig Question

Hello, D!ddy!

Subhotosh provided an excellent solution.
. . Here's an alternate approach.

If \(\displaystyle \cos\theta \:=\:\text{-}\sin\theta\), then \(\displaystyle \sin^2\!\theta \:=\:?\)

. . \(\displaystyle (A)\; \frac{1}{4}\qquad(b)\;\frac{1}{2} \qquad (C)\;\frac{\sqrt{2}}{2} \qquad (D)\;-\frac{\sqrt{2}}{2}\)

Square both sides: .\(\displaystyle \cos^2\!\theta \:=\:\sin^2\!\theta \quad\Rightarrow\quad 1-\sin^2\!\theta \:=\: \sin^2\!\theta\quad\Rightarrow\quad 2\sin^2\!\theta \:=\:1\)

\(\displaystyle \text{Therefore: }\;\sin^2\!\theta \:=\:\frac{1}{2}\quad\hdots\;\text{ answer (b)}\)