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How do you use the sum or difference identity to find the excat value of sin 165°​

Sagot :

Answer:

[tex]\frac{1}{4}[/tex] ([tex]\sqrt{6}[/tex] - [tex]\sqrt{2}[/tex] )

Step-by-step explanation:

Using the addition formula for sine

sin(A + B) = sinAcosB + cosAsinB

Given

sin165° = sin(120 + 45)° , then

sin(120 + 45)°

= sin120°cos45° + cos120°sin45°

= sin60°cos45°  - cos60°sin45°

= [tex]\frac{\sqrt{3} }{2}[/tex] × [tex]\frac{\sqrt{2} }{2}[/tex] - [tex]\frac{1}{2}[/tex] × [tex]\frac{\sqrt{2} }{2}[/tex]

= [tex]\frac{\sqrt{6} }{4}[/tex] - [tex]\frac{\sqrt{2} }{4}[/tex]

= [tex]\frac{1}{4}[/tex] ( [tex]\sqrt{6}[/tex] - [tex]\sqrt{2}[/tex] )