How do you use the sum and difference identities to find the exact value of ${cos 15}^{\circ}$ $cos 15^@$ ?

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Answer 1 · 2014-10-24T04:22:09

The special triangles, 30-60-90 and 45-45-90, allow us to evaluate sin and cos.

We leverage that information to evaluate $cos (15)$ $cos(15)$ .

$cos (a - b) = cos (a) cos (b) + sin (a) sin (b)$ $cos(a-b)=cos(a)cos(b)+sin(a)sin(b)$

$cos (60 - 45) = cos (60) cos (45) + sin (60) sin (45)$ $cos(60-45)=cos(60)cos(45)+sin(60)sin(45)$

$cos (15) = cos (60) cos (45) + sin (60) sin (45)$ $cos(15)=cos(60)cos(45)+sin(60)sin(45)$

$cos (15) = \frac{1}{2} \cdot \frac{1}{\sqrt{2}} + \frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{2}}$ $cos(15)=1/2*1/sqrt(2)+sqrt(3)/2*1/sqrt(2)$

$cos (15) = \frac{1}{2 \sqrt{2}} + \frac{\sqrt{3}}{2 \sqrt{2}}$ $cos(15)=1/(2sqrt(2))+sqrt(3)/(2sqrt(2))$

$cos (15) = \frac{1 + \sqrt{3}}{2 \sqrt{2}}$ $cos(15)=(1+sqrt(3))/(2sqrt(2))$

$cos (15) = \frac{1 + \sqrt{3}}{2 \sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}$ $cos(15)=(1+sqrt(3))/(2sqrt(2))*sqrt(2)/sqrt(2)$

$cos (15) = \frac{\sqrt{2} + \sqrt{6}}{4}$ $cos(15)=(sqrt(2)+sqrt(6))/4$

$cos (15) = 0.9659258263$ $cos(15)=0.9659258263$

Verify your results using a calculator. Make sure the calculator is in Degree Mode.

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How do you use the sum and difference identities to find the exact value of cos15∘cos 15^@?