Question

If A = 15° , find the value of cos² 3A + 4cos4A – sin6A.

Answer 1

Step-by-step explanation:

$\cos {}^{2} (45) + 4 \cos(60) \\ = ( \frac{1}{ \sqrt{2} } ) {}^{2} +4 \times \frac{1}{2} \\ = \frac{1}{2} + 2 = \frac{5}{2}$

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Answer 2

given★

cos² 3A + 4cos4A – sin6A.------(1

where A=15°---------(2

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solution★

EQ (1 value put on EQ (2

$= \cos {}^{2} (3 \times 15) + 4 \cos(4 \times 15) - \sin(6 \times 15) \\ \\ = \cos {}^{2} (45) + 4 \cos(60) - \sin(90) ..(3$

we know that

$= > \cos(45) = \frac{1}{ \sqrt{2} } \\ \\ = > \cos(60) = \frac{1}{2} \\ \\ = > \sin(90) = 1....(4$

EQ 4 value put on EQ 3

$=(\frac{1}{ \sqrt{2} }) {}^{2} + 4 \times \frac{1}{2} - 6 \times 1 \\ = \frac{1}{2} + 2 - 6 \\ \\ = \frac{1}{2} - 4 \\ \\ = \frac{ - 7}{2}$

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I hope it helps you