Question

sin° (1 + ‌‌sin2°) = cos 2° then prove that cos6°-4 cos4° + 8 cos2°=4​

Answer 1

Let us first assume that,

1 + sin(2) = t

Now we have the first case as,

sin(t) = cos(2)

Now, we have

t = $sin^{-1}$[cos(2)]

t = $sin^{-1}$[sin(90-2)] = 88 degree.

Now, cos(6) - 4[cos(4)] + 8[cos(2)]

cos[2(3)] - 4[cos{2(2)} + 8 cos(2)

2 $cos^{2}$(3) - 1 -4[ 2$cos^{2}$(2) -1] + 8 cos(2)

We used the formula, cos(2A) = 2$cos^{2}$A - 1

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I have come this far in the question. Here, even the calculator gives a different answer.

Either you have given wrong values in the question, or, $(sin)^{0}$(x) does not exist, only sin$(x)^{0}$ exists, check it carefully.

So please correct the question.