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

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

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...

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

tana+tanb/1-tanatanb im confused becase sqrt of 3 over 3 is the value of tan 30 so its hard to solve the arithmetic and the algebra. help please!

I chose to manipulate the left hand side. So i got 1+cos2x-cos2x-cos4x when i simplify it will be 1-cos4x i will break it into pieces so (1-cos2x)(1+cos2x) i can change the first term...

If there is no solution, enter NO SOLUTION.) 2 -2 sin t = 2 sqrt(3)cos t My solutions were pi/2, 7pi/6, and 11pi/6.

Please help, I have no idea! Solve the following trigonometric equation sin^2x - cos^2x = 1/4.

(secθ-tanθ)^2=2secθtanθ

I can find x = 0, and also can show that cos x = (1/4)*(-1+/-sqrt(5)), but I don't know how I should have known that the latter cosines of x correspond to angles involving fifths of pi (radians)...

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