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Cos(A+B)Cos(A-B)

sin 4A = cosA...

Question

$sin 4 A = cos (A - 20^{o})$ , where 4A is an acute angle, find the value of A in degrees

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Solution

We know that,
$sin A = cos (90^{o} - A)$

Now, $sin 4 A = c o s (A - 20^{o}) \Rightarrow cos (90^{o} - 4 A) = c o s (A - 20^{o}) \Rightarrow 90^{o} - 4 A = A - 20^{o} \Rightarrow 110^{o} = 5 A \Rightarrow A = 22^{o}$

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Cos(A+B)Cos(A-B)

Standard XII Mathematics

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