If $\sin(2x) = \frac{3}{5}$
Find $\cos(4x)$..
I tried by : $\cos(4x)= \cos(2\cdot2x)$..
And $\cos (2\cdot2x) = 1-2\sin^2(2x)$ ..
From it ---- $\cos(4x)=0.28$.
Is there any other ways ?
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2That's the most direct one, in my opinion. – Apr 13 '17 at 03:54
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2Looks fine to me since the answer is $\frac{7}{25}$ which is $0.28$ (except your formatting)--) – Juniven Acapulco Apr 13 '17 at 03:54
2 Answers
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Form the 3-4-5 triangle.
Reflect it about the "4" side. to get a 5-5-6 triangle.
apply the cosine law to it.
Mick
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Your method seems the 'obvious' way. However you asked for other ways so here is another.
Given $\sin2x=3/5$ you can see from the right angled triangle with opposite $= 3$, hypotenuse $= 5$ that the adjacent side $= 4$ and so $\cos2x=4/5$
Knowing $\sin2x$ and $\cos2x$ you can determine $\sin4x=2\sin2x\cos2x=24/25$
These values are the opposite and hypotenuse of another nice triangle whose adjacent side $=7$. Hence $\cos4x=7/25=0.28$
PM.
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