Trigonometry help?

Question

11. Assume that sin⁡u = 2/3 and sinv = 1/3 and that u and v are between 0 and π/2. Evaluate sin(u+v).



12. Prove the following identity: ( sin⁡2θ / sin⁡θ ) - ( cos⁡2θ / cos⁡θ ) = sec⁡θ

 

Answer 1

11).

Since u and v are between 0 and π/2.this means that, they are both in quadrant 1.

If sin u = 2/3, then cos u = √(5)/3                             (since, cos u = √[ 1 - sin² (u) ])

If sin v = 1/3, then cos v = 2√(2)/3                            (since, cos v = √[ 1 - sin² (v) ])

⇒ sin (u + v) = sin u cos v + cos u sin v.

Substitute the values sin u = 2/3, sin v = 1/3, cos u = √(5)/3, and cos v = 2√(2)/3 in

sin (u + v) = sin u cos v + cos u sin v.

sin (u + v) = (2/3)(2√(2)/3) + (√(5)/3)(1/3)

= (4√2/9) + (√(5)/9)

= (4√2 + √5)/9.

Therefore, sin (u + v) = (4√2 + √5)/9.

Comments

The sum formula : sin(u + v) = sin(u) cos(v) + cos(u) sin(v).

The trigonometric values are sin(⁡u) = 2/3 and sin(v) = 1/3 and that u and v are lies between 0 and π/2.

From Pythagorean identity : cos(u) = √[ 1 - sin²(u) ]

cos(u) = √[ 1 - (2/3)² ] = √[ 1 - 4/9 ] = √[ (9 - 4)/9 ] = √[ 5/9 ] =  √5/3.

From Pythagorean identity : cos(v) = √[ 1 - sin²(v) ]

cos(v) = √[ 1 - (1/3)² ] = √[ 1 - 1/9 ] = √[ (9 - 1)/9 ] = √[ 8/9 ] =  √8/3.

sin(u + v) = sin(u) cos(v) + cos(u) sin(v)

sin(u + v) = (2/3) (√8/3) + (√5/3) (1/3)

sin(u + v) = 2√8/9 + √5/9

sin(u + v) = (4√2 + √5 )/9

Answer 2

12).

The trigonometric equation is (sin 2θ / sinθ) - (cos 2θ / cos θ) = sec θ.

The left hand side identity is (sin 2θ / sinθ) - (cos 2θ / cos θ)

= (2sin θ cos θ/sin θ) - [ (2cos² θ - 1)/cos θ ]

= 2cos θ - [ (2cos² θ - 1)/cos θ ]

= (2cos² θ -  2cos² θ + 1)/cos θ

= 1/cos θ

= sec θ

= right hand side identity.