Srry to take up ur guys' time like this but I don't know how to do these questions and my friends don't know how to do it either so can someone please show me the thought process behind these??


1. given that cos(2theta)=5/42 --find product of all possible values of sin(theta)

--for this question do you solve the first one and then plug it in??? I'm so confused is there a technique to do this.. or??


2. if the angle theta is-- 2pi ≤ theta ≤ 4pi and cos(theta)= -7/25, what is sin(theta/2)

-- I actually tried this one and tried drawing a triangle but then I got a quadratic?? is that right


3. if 1+cos^2(theta)+cos^4(theta)+cos^6(theta)+... =5  what is the value of cos(2theta)

-- I don't even know how to tackle this problem crying

 Nov 27, 2022

I'm so dearly sorry !! I just read the guidelines saying one question per post.. I won't repeat the same mistake next time-- though could I seek help on these just this once?? I'm so confused

 Nov 27, 2022

Hi fluufye!

Question 1:

Given: \(cos(2\theta)=\dfrac{5}{42}\), find the product of all possible values of \(sin(\theta)\)

Well, for this type of question, you need to recall the formula: \(sin^2(\theta)=\dfrac{1-cos(2\theta)}{2}\) (Half-angle identity).



\(\implies sin(\theta)=\pm \dfrac{\sqrt{777}}{42}\), so \(sin(\theta)\) has two values, and their product is:

\(\dfrac{\sqrt{777}}{42}*(-\dfrac{\sqrt{777}}{42})=-\dfrac{37}{84}\) which should be the answer.

Again, for the next question, question 2:
We need the half-angle identity again! Remember: \(sin^2(x)=\dfrac{1-cos(2x)}{2}\), now let \(x=\dfrac{\theta}{2}\)

We get: \(sin^2(\dfrac{\theta}{2})=\dfrac{1-cos(\theta)}{2}\)

We are given that \(cos(\theta)=-\dfrac{7}{25}\) 



So, taking the square root, we get:

\(sin(\frac{\theta}{2})=\pm \dfrac{4}{5}\)

But is it negative or positive or both answers are correct?? We must refer to the domain for this issue.

We are given that, \(2\pi \le \theta \le 4\pi\)

Now divide this inequality by 2:

\(\pi \le \dfrac{\theta}{2} \le 2\pi\)

What is this quadrant? This starts from the third quadrant and till the end of the fourth quadrant. I.e. theta/2 could be either in the third or fourth quadrant. But it does not matter which, because in either case, sin(theta/2) is negative (Recall sin is negative in both third and fourth quadrants.)

Therefore, we reject the positive answer, and we only choose the negative answer, giving: \(sin(\frac{\theta}{2})=-\dfrac{4}{5}\)


Lastly, for the third question: I'll use x instead of theta.

\(1+cos^2(x)+cos^4(x)+cos^6(x)+...=5\). The questions wants the value of \(cos(2x)\)

A good way to start this question is observing that the given sum is a geometric series.

The first term is 1, then the second is cosine squared, then the third is cosine to the power of 4 (I.e. the square of the previous term.)
I notice that to get the next term, we multiply by cos squared.

So the common ratio, and the first term are : \(r=cos^2(x) \\ a=1\) respectively.

Remember the geometric series sum formula: \(S=\frac{a}{1-r}, \text{ where } \left | r\right | <1\)

The condition is already satisfied.

So we are given that S = 5 and we have a and r, by substituting we get:


Let's solve for cosine squared.

\(1-cos^2(x)=\dfrac{1}{5} \\ \iff cos^2(x)=\dfrac{4}{5}\)

But the question wants cos(2x).....

Again, another identity. (The double angle identity) \(cos(2x)=cos^2(x)-sin^2(x)=2cos^2(x)-1=1-2sin^2(x)\)

Are all equivalent. However, we pick the fastest one here, namely: \(cos(2x)=2cos^2(x)-1\), because we already have cosine squared!

So: \(cos(2x)=2(\dfrac{4}{5})-1=\dfrac{8}{5}-1=\dfrac{3}{5}\) which is the answer.



I hope this helps, and do not hesitate to ask for further explanations.

 Nov 27, 2022

Ohhhh so you had to use the half-angle identity equation-- thank you so much haha!! I'll try and attempt the question again knowing the formula... thank you so much once again!! Also I understood the concepts on the 2nd and 3rd question on where I went wrong and how to do the question it self. this is really helpful!!

fluufye  Nov 28, 2022
edited by fluufye  Nov 28, 2022

you are welcome!

Guest Nov 28, 2022

7 Online Users