Question

prove that cos^2x=sinx
​

Answer 1

Answer:

cos2x =

cos

2

x

−

sin

2

x

SO,

cos

2

x

−

sin

2

x

- sinx =0

1-

sin

2

x

-

sin

2

x

-sinx=0

1-2

sin

2

x

-sinx=0

2

sin

2

x

+sinx -1 =0

2

sin

2

x

+(2-1) sinx -1=0

2

sin

2

x

+2sinx - sinx -1 =0

2sinx(sinx +1) -1(sinx +1)=0

(sinx+1) ( 2sinx -1) =0

either or

sinx=-1 sinx=1/2

So you can solve for these values of sinx.

Answer 2

Step-by-step explanation:

TRIGONOMETRY PROOFS

Caleb S. asked • 09/28/17

Prove that Cos(2x) = -Sin(x)

Took a quiz in precal and this question completely stumped me to the point of writing it down to try and figure it out later. I know you would start with the left side but that’s as far as I have gotten.

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1 Expert Answer

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Andy C. answered • 09/28/17

TUTOR 4.9 (27)

Math/Physics Tutor

SEE TUTORS LIKE THIS

for x=0, cos(2*0) =1 while -sin(0) = 0

for x=pi, cos(2*pi) = 1 while -sin(pi) = 0

for x = pi/2, the cosine is 1 while the sine is -1

So no, they are not EXACTLY the same.

--------------------------------------------------------------

But we can find where they are equal:

cos(2x) = - sin(x)

1 - 2*(sinX)^2 = -sinX

Let z = sinX

1 - 2z^2 = -z

1 + z - 2z^2 = 0

2z^2 - z - 1 = 0

( 2z + 1)( z - 1 ) = 0

-----------------------------------

2z + 1 = 0 ---> z = -1/2

sin(x) = -1/2

x = 210, 330

x = (7/6)*pi, or x = (11/6)*pi

-------------------------------------

z - 1 = 0 ---> z =1

sin(x) = 1

x = 90 ---> x = (pi/2)