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# C3 trig watch

1. I have to put the equation in the photo in terms of one trigonometric function. What is wrong with what I did? The answer is meant to be 1/4sin^2(2x)

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2. What you done there is correct. Show what the actual question was.
3. Rotate that picture, or use latex (google latex tsr and it should come up to the thread, I can't remember where it is located)

I'm struggling to understand what you want too
4. (Original post by ubisoft)
What you done there is correct. Show what the actual question was.
It's 3f

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5. (Original post by JD1lla)
Rotate that picture, or use latex (google latex tsr and it should come up to the thread, I can't remember where it is located)

I'm struggling to understand what you want too
I am using no the double angle formula: sin2A = 2sinAcosA

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6. (Original post by anoymous1111)
I have to put the equation in the photo in terms of one trigonometric function. What is wrong with what I did? The answer is meant to be 1/4sin^2(2x)

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I get the same answer like you in 2 different ways
7. (Original post by JD1lla)
Rotate that picture, or use latex (google latex tsr and it should come up to the thread, I can't remember where it is located)

I'm struggling to understand what you want too

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8. (Original post by TeeEm)
I get the same answer like you in 2 different ways
Apparently it is wrong.... I don't know why

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9. (Original post by anoymous1111)
It's 3f

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Why did you just square root it out of nowhere? that changes the expression.

Think of double angle expressions that you can use.
10. (Original post by anoymous1111)
Apparently it is wrong.... I don't know why

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Post your methodology, I don't know how you're using that identity (well I can guess). I don't know how the posters above have also got the same answer as you. Are you all mis-reading the question/OPs post?

She wrote cos^2, not cos2
11. (Original post by JD1lla)
Post your methodology, I don't know how you're using that identity (well I can guess). I don't know how the posters above have also got the same answer as you. Are you all mis-reading the question/OPs post?

She wrote cos^2, not cos2
There isn't really any methodology....like obviously 1/2(2) =1 so I'm just rewriting the equation so that I can use the formula to simplify part of it....

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12. (Original post by anoymous1111)
There isn't really any methodology....like obviously 1/2(2) =1 so I'm just rewriting the equation so that I can use the formula to simplify part of it....

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But you can't square root something that isn't part of an equation
You can rewrite it as
(sinthetacostheta)^2 and then proceed with the method you did and expand at the end.
13. (Original post by anoymous1111)
Apparently it is wrong.... I don't know why

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it is wrong because now that you posted the question there is no square root anywhere in the question....
14. (Original post by anoymous1111)
There isn't really any methodology....like obviously 1/2(2) =1 so I'm just rewriting the equation so that I can use the formula to simplify part of it....

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Square rooting the whole thing is not rewriting it; it changes the expression. It's not an equation.
15. (Original post by TeeEm)
it is wrong because now that you posted the question there is no square root anywhere in the question....
Ooooooooooops sorry everyone I'm so stupid.....I thought I had included the question. Why can't I square root the whole thing? Is it because it's not stated that it is equal to 0 therefore I can only rearrange it

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16. is what I got, which is another form of the answer you seek, but I assume I used different methodology to the way you require. I suppose this doesn't help :P
17. (Original post by anoymous1111)
Ooooooooooops sorry everyone I'm so stupid.....I thought I had included the question. Why can't I square root the whole thing? Is it because it's not stated that it is equal to 0 therefore I can only rearrange it

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If you have the value 9.
9 is identical to 3^2 but it isn't identical to sqrt(9)
Just like
cos^2sin^2 is identical to (costhetasintheta)^2 but not identical to sqrt(cos^2thetasin^2theta)
18. (Original post by anoymous1111)
I have to put the equation in the photo in terms of one trigonometric function. What is wrong with what I did? The answer is meant to be 1/4sin^2(2x)

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Factorise it first into (sinxcosx)^2, and then do what you did for inside the bracket and then expand it (except you do not square root it).
19. (Original post by Windowswind123)
If you have the value 9.
9 is identical to 3^2 but it isn't identical to sqrt(9)
Just like
cos^2sin^2 is identical to (costhetasintheta)^2 but not identical to sqrt(cos^2thetasin^2theta)
Haha I know I don't know what I was thinking! Sorry I wasted your time!

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20. (Original post by JD1lla)
is what I got, which is another form of the answer you seek, but I assume I used different methodology to the way you require. I suppose this doesn't help :P
That's great thank you

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