Questions LLC
Login
or
Sign Up
Ask a New Question
Mathematics
Trigonometry
Solving Trigonometric Equations
Did you solve sin2[sin^-1(pi/6)] and sin[2sin^-1(pi/6)] in the same way?
1 answer
geez - stop starting new threads for follow-up questions.
You can
ask a new question
or
answer this question
.
Related Questions
Write the equation for the graph of the rose curve.
(1 point) Responses r = 4 sin 9θ r = 4 sin 9 θ r = 8 sin 9θ r = 8 sin 9 θ
4. Asked to simplify the expression sin(180−è), Rory volunteered the following solution:
sin(180−è) = sin 180−sin è,
Which of the following are true statements? Select all that apply. (1 point) Responses sin 67° = sin 42° cos 25° + cos 42°
If A + B + C = 180°, Prove that
Sin² (A/2) + sin² (B/2) + sin²(C/2) = 1 – 2sin (A/2) sin (B/2) sin(C/2)
d/dx( ln |sin(pi/x)| ) = ?
Thanks. If those are absolute value signs, the derivative will not exist when sin (pi/x) = 0, because
I need to find the exact solutions on the interval [0,2pi) for:
2sin^2(x/2) - 3sin(x/2) + 1 = 0 I would start:
Verify the identities.
Cos^2x - sin^2x = 2cos^2x - 1 When verifying identities, can I work on both side? Ex. 1 - sin^2x - sin^2x
If sin(θ) = -1/√2, find sin2(θ). (Recall that sin2(θ) means (sin(θ))2.)
sin2(θ) =
Add and subtract as indicated. Then simplify your answer if possible. Leave answer in terms of sin(θ) and/or cos(θ). (Remember
1. Solve: 2 cos² x - 3 cos x + 1 = 0 for 0 ≤ x < 2pi.
2. Solve: 2 sin x - 1 = 0 for 0° ≤ x < 360° 3. Solve: sin² x =
