Solve step by step to get the answer

Solve step by step to get the answer

The question is- In a △PQR, if 3sinP+4cosQ = 6 and 4sinQ+3cosP=1, then the angle R is equal to:

sinxcos^4x+cos^6x=

Hi, this is my first time using this site. So I have been staring at this trigonometric equation for a long while and looking for what it asks for. For what I know, I think I have to use sum of...

Prove sin4x = 4 (sin x-2 sin^3 x)/secx I can't seem to solve this! Halp!!!

Solve the given equation over the interval [0,2): 2sinx+ (sqrt 3) sin^2x=-(sqrt 3) cos^2x A. x=0, and x=(pi/2), and x=(3 pi/2), and x=2(pi)x=(3 pi/2), and x=(7 pi/2) b. x=(4 pi/3) and...

Provide an exact evaluation of the expression: tan (-285°).

Find the value for sin (θ) if the following conditions hold: cos(2(theta)) = 3/5 and 90(degrees) < (theta) < 180(degrees) A. (1/25) B. (sqrt5/25) C. (2 sqrt5/5) D. (sqrt5/5)

Sin(45-x)=cos(45+x) How to prove this identity?

I thought it'd be (1 / cos(θ)2) * (1 / tan(θ)) * (cos(θ) / 1) Which would get you cos(θ) / cos(θ)2 * tan(θ) And then coming out to 1 / cos(θ) *...

cos7y cos3y -sin 7y sin3y= ?

i need step by step soln plz thank u

Please help me solve this equation to better help me to understand trig Identites.

I am a little confused on this and it would be nice if someone could work through it so I can understand this problem

1) (cscx + cotx) / (tanx + sinx) = cotxcscx 2) (1/secx) -secx = -sinx 3) (1-tan^2x / 1-cot^2x) = -tan^2x

could someone explain finding cos(11pi/12) using Sum and difference Identities. it must use one of these: cos(α+β)=cos(α)cos(β)−sin(α)sin(β) or cos(α+β)=cos(α)cos(β)+sin(α)sin(β) Im...

=(sinx/cosx - sinx)/ (2sinx/cosx) = sinx-sinxcosx/cosx • cosx/2sinx = sinx (1-cosx)/2sinx = 1-cosx/2 now im lost help!!

Using only the double angle identities how to solve for this? im stucked after 2cos^2 (2x) -1

I know that i should multiply something to it for it to become 2sinAcosA for me to be able to solve it. Unfortunately, if i multiply2/2 to it will just become 2/16 that will just bounce back to 1/8...

tan34/2(1-tan34) =(2/2)[tan34/2(1-tan34) =2tan34/4(1-tan34) i need to have the form 2tanA/1-tan^2A but i cant eliminate the 4

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