cos a = ± 1 − s i n 2 a , tan a = sin a / cos a \cos a = \pm \sqrt{1 - sin^2 a}, \tan a = \sin a / \cos a cos a = ± 1 − s i n 2 a , tan a = sin a / cos a
Precalculate:
cos x = 1 − ( 4 / 5 ) 2 = 3 / 5 , tan x = 4 / 3 \cos x = \sqrt{1 - (4/5)^2} = 3/5, \tan x = 4/3 cos x = 1 − ( 4/5 ) 2 = 3/5 , tan x = 4/3
cos y = 1 − ( − 12 / 13 ) 2 = 5 / 13 , tan y = − 12 / 5 \cos y = \sqrt{1 - (-12/13)^2} = 5/13, \tan y = -12/5 cos y = 1 − ( − 12/13 ) 2 = 5/13 , tan y = − 12/5
Answer:
a ) cos ( x + y ) = cos x ⋅ cos y − sin x ⋅ sin y = 3 / 5 ⋅ 5 / 13 − 4 / 5 ⋅ ( − 12 / 13 ) = 63 / 65 \begin{aligned}a) \cos(x+y) &= \cos x \cdot \cos y - \sin x \cdot \sin y \\
&= 3/5 \cdot 5/13 - 4/5 \cdot (-12/13) \\
&= 63/65\end{aligned} a ) cos ( x + y ) = cos x ⋅ cos y − sin x ⋅ sin y = 3/5 ⋅ 5/13 − 4/5 ⋅ ( − 12/13 ) = 63/65
b ) tan ( x − y ) = ( tan x − tan y ) / ( 1 + tan x ⋅ tan y ) = ( 4 / 3 − ( − 12 / 5 ) ) / ( 1 + 4 / 3 ⋅ ( − 12 / 5 ) ) = 56 / ( 15 − 48 ) = − 56 / 33 \begin{aligned} b) \tan(x-y) &= (\tan x - \tan y) / (1 + \tan x \cdot \tan y) \\
&= (4/3 - (-12/5)) / (1 + 4/3 \cdot (-12/5)) \\
&= 56 / (15 - 48 ) = -56 / 33\end{aligned} b ) tan ( x − y ) = ( tan x − tan y ) / ( 1 + tan x ⋅ tan y ) = ( 4/3 − ( − 12/5 )) / ( 1 + 4/3 ⋅ ( − 12/5 )) = 56/ ( 15 − 48 ) = − 56/33
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