10. By using the result for sin2α
into our RHS and obtain:
(remember:
example
11. Change sin 70◦cos 150 ◦+cos70 ◦sin150◦into a
trigonometric function in a single variable and
evaluate it.
Answer:
This is one side of sum idnetity for sines :
sin(α+β)=sin α.cos β+cos α.sin β
sin 70◦cos 150 ◦+cos70 ◦sin150◦ = sin (70◦+150◦ )
= sin (220◦)
= -sin 220◦
=-sin (220 ◦ - 180 ◦)
=-sin 40 ◦
=-.643
180 ◦ <220 ◦ <270 ◦
Quadrant lll
=-cos(270-220)
=-cos 50
=-.643
12. Change sin 60◦cos 45 ◦- cos 60 ◦sin45◦into a
trigonometric function in a single variable and
evaluate it.
Answer:
This is part of the difference identity for sines :
sin(α-β)=sin α.cos β-cos α.sin β
sin 60◦cos 45 ◦- cos 60 ◦sin150◦ = sin (60◦-45◦ )
= sin (15◦)
= sin 15 ◦
=.259
13. Change cos 85◦cos 15 ◦- sin85◦ sin15◦into a
trigonometric function in a single variable and
evaluate it.
Answer:
This is part of the sum identity for cosines :
cos(α+β) = cos α.cos β – sin α.sin β
cos 85◦cos 15 ◦- sin85◦ sin15◦ = cos (85◦+15◦ )
= cos (100◦)
= -cos 100 ◦
=-cos(180 ◦ - 100◦)
=-cos 80 ◦
=-.087
14. 4
5
3
α
If P in the second quadrant and sinP= , find sin2P.
Answer:
If sinP= , then cosP=
sin2P = 2sinP.cosP
sin2P= 2
=
15. If A is a second quadrant angle, and sinA = what is cos2A ?
Answaer:
Use pythagorean tripels : (5, 12, 13)
If sinA= , then cosA=
If we want to use the formula of :
Cause cosine in negative in
quadrant II
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