Questions LLC
Login
or
Sign Up
Ask a New Question
Mathematics
Trigonometry
Trigonometric Identities
Solve any way possibly? Thank you!
1-sin^2x/1-cosx
1 answer
2sinxcosx/1-cos =sinxcosx/1-cos=-sinx
You can
ask a new question
or
answer this question
.
Similar Questions
I can't seem to prove these trig identities and would really appreciate help:
1. cosx + 1/sin^3x = cscx/1 - cosx I changed the 1:
Top answer:
(cos+1)/sin^3 = csc/(1-cos) multiply by (1-cos) (1+cos)(1-cos)/sin^3 = csc = 1/sin (1-cos^2)/sin^2 =
Read more.
Solve:
cos(2x-180) - sin(x-90)=0 my work: cos2xcos180 + sin2xsin180= sinxcos90 - sin90cosx -cos2x - sin2x= cosx -cos^2x + sin^2x
Top answer:
To solve the equation cos(2x-180) - sin(x-90) = 0, let's break it down step by step. 1. Rewrite the
Read more.
Verify the identity:
sin^(1/2)x*cosx - sin^(5/2)*cosx = cos^3x sq root sin x I honestly have no clue how to approach the
Top answer:
since 5/2 = 2 + 1/2, you have u^5/2 = u^2 * u^1/2, and so, √sinx cosx - sin^2x √sinx cosx
Read more.
1. Create an algebraic expression for
sin(arccosx-arcsin3x) 2. The cosx=4/5, x lies in quadrant 4. Find sin x/2 3.Determine all
Top answer:
Wow! You actually want somebody to do that assignment for you, without showing any effort on your
Read more.
hey, i would really appreciate some help solving for x when:
sin2x=cosx Use the identity sin 2A = 2sinAcosA so: sin 2x = cos x
Top answer:
To solve the equation sin 2x = cos x, we can use the identity sin 2A = 2 sin A cos A. This allows us
Read more.
Solve this equation fo rx in the interval 0<=x<=360
3sinxtanx=8 I would do it this way: sinxtanx = 8/3 sinx(sinx/cosx)=8/3
Top answer:
To solve the equation 3sinxtanx = 8 for x in the interval 0 <= x <= 360, follow these steps: 1.
Read more.
Simplify #3:
[cosx-sin(90-x)sinx]/[cosx-cos(180-x)tanx] = [cosx-(sin90cosx-cos90sinx)sinx]/[cosx-(cos180cosx+sinx180sinx)tanx] =
Top answer:
looks good to me
Read more.
Prove each idenity.
1+1/tan^2x=1/sin^2x 1/cosx-cosx=sinxtanx 1/sin^2x+1/cos^2x=1/sin^2xcos^2x 1/1-cos^2x+/1+cosx=2/sin^2x and
Top answer:
Sure, I'll walk you through the steps to prove each of the identities. Let's start with the first
Read more.
I'm sorry to double post; I don't want to seem impatient, but I really need help with this.
Prove each idenity.
Top answer:
i will do the first one 1+1/tan^2x=1/sin^2x LS = 1 + 1/(sin^2x/cos^2x) = 1 + cos^2x/sin^2x = (sin^2x
Read more.
solve each equation for 0=/<x=/<2pi
sin^2x + 5sinx + 6 = 0? how do i factor this and solve? 2sin^2 + sinx = 0 (2sinx - 3)(sinx
Top answer:
sin^2x + 5sinx + 6 = 0? how do i factor this and solve? by saying y = sin x then you have y^2 + 5 y
Read more.
Related Questions
How do I solve this?
tan^2x= 2tanxsinx My work so far: tan^2x - 2tanxsinx=0 tanx(tanx - 2sinx)=0 Then the solutions are: TanX=0
Suppose f(x) = sin(pi*cosx) On any interval where the inverse function y = f –1(x) exists, the derivative of f –1(x) with
How many solutions does the equation cosx + 1/2 = 1 have for 0<x<2pi
cosx+1/2=1 cosx+1=2 cosx=2-1 cosx=1 therefore 1 solution? A)
help me solve this coz im stumped
secx-secx sin^2x= cosx thanks
Suppose f(x) = sin(pi*cosx) On any interval where the inverse function y = f ^–1(x) exists, the derivative of f ^–1(x) with
(Sin^3x-cos^3x)/(sinx-cosx) – cosx/sqrt(1+cot^2x)-2tanxcotx=-1 where x∈(0,2pi)
general value of x.
solve 4 sin^2x + 4 sqrt 2 cos x-6=0 for all real values of x.
Going by the example in the book I got to: 4 cos^2x + 4sqrt2 cosx
Write the equation for the function based on the graph.
(1 point) Responses y=cosx y=cosx y = cos(x) −2 y = cos(x) −2 y =
If secx = 8 and -pi/2 < x < 0, find the exact value of sin2x
Use the identity sin 2x = 2(sinx)(cosx) if secx = 8, then cosx = 1/8
prove the following trig identity:
a) sin(pi/6 + x) + sin (pi/3 + x) + sin (pi/2 + x) = ((sqrt3) +1)/2 (sinx +(sqrt3)cosx) b)