Question

use the sum/difference identities to evaluate exactly
cos 150 degree

Answer 1

\[\cos 150^0 = \cos(90^0 + 60^0)\]
\[= \cos(90^0 + 60^0) = \cos90^0 \cos 60^0 - \sin90^0 \sin60^0\]
\[= 0 \times \frac{1}{2} - 1 \times \frac{\sqrt 3}{2} = 0 - \frac{\sqrt 3}{2}\]
\[= - \frac{\sqrt 3}{2}\]
Hence; \[\cos 150^0 =- \frac{\sqrt 3}{2}\]

Formula used
cos(A+B) = cos A cos B- sin A sin B
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Answer 2

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Answer 3

why i cant use sin(45-30)

Answer 4

never mind

Answer 5

Sure but you wiil have to convert cos 150 into sin

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Answer 6

cos105 degree

Answer 7

\[\cos150^0=\cos(180^0-30^0)= -\cos30^0 = - \frac {\sqrt3}{2}\]

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Answer 8

Another way:

Answer 9

the question is cos 105 degree

Answer 10

\[\cos150^0=\cos(90^0+60^0)= -\sin60^0 = - \frac {\sqrt3}{2}\]

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Answer 11

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Answer 12

i got the wrong question