Answer:
sin⁴x - sin²x = cos⁴x - cos²x
Solve the right hand side of the equation
That's
sin⁴x - sin²x
<u>From trigonometric identities</u>
<h3>
sin²x = 1 - cos²x</h3>
So we have
sin⁴x - ( 1 - cos²x)
sin⁴x - 1 + cos²x
sin⁴x = ( sin²x)(sin²x)
That is
( sin²x)(sin²x)
So we have
( 1 - cos²x)(1 - cos²x) - 1 + cos²x
<u>Expand</u>
1 - cos²x - cos²x + cos⁴x - 1 + cos²x
1 - 2cos²x + cos⁴x - 1 + cos²x
<u>Group like terms</u>
That's
cos⁴x - 2cos²x + cos²x + 1 - 1
<u>Simplify</u>
We have the final answer as
<h3>cos⁴x - cos²x</h3>
So we have
<h3>cos⁴x - cos²x = cos⁴x - cos²x</h3>
Since the right hand side is equal to the left hand side the identity is true
Hope this helps you
Answer:
The mid-point between the endpoints (10,5) and (6,9) is:
Step-by-step explanation:
Let (x, y) be the mid-point
Given the points
Using the formula to find the mid-point between the endpoints (10,5) and (6,9)
![\left(x,\:y\right)=\left(\frac{x_2+x_1}{2},\:\:\frac{y_2+y_1}{2}\right)](https://tex.z-dn.net/?f=%5Cleft%28x%2C%5C%3Ay%5Cright%29%3D%5Cleft%28%5Cfrac%7Bx_2%2Bx_1%7D%7B2%7D%2C%5C%3A%5C%3A%5Cfrac%7By_2%2By_1%7D%7B2%7D%5Cright%29)
Here:
![\left(x_1,\:y_1\right)=\left(10,\:5\right),\:\left(x_2,\:y_2\right)=\left(6,\:9\right)](https://tex.z-dn.net/?f=%5Cleft%28x_1%2C%5C%3Ay_1%5Cright%29%3D%5Cleft%2810%2C%5C%3A5%5Cright%29%2C%5C%3A%5Cleft%28x_2%2C%5C%3Ay_2%5Cright%29%3D%5Cleft%286%2C%5C%3A9%5Cright%29)
Thus,
![\left(x,\:y\right)=\left(\frac{x_2+x_1}{2},\:\:\frac{y_2+y_1}{2}\right)](https://tex.z-dn.net/?f=%5Cleft%28x%2C%5C%3Ay%5Cright%29%3D%5Cleft%28%5Cfrac%7Bx_2%2Bx_1%7D%7B2%7D%2C%5C%3A%5C%3A%5Cfrac%7By_2%2By_1%7D%7B2%7D%5Cright%29)
![\left(x,\:y\right)=\left(\frac{6+10}{2},\:\frac{9+5}{2}\right)](https://tex.z-dn.net/?f=%5Cleft%28x%2C%5C%3Ay%5Cright%29%3D%5Cleft%28%5Cfrac%7B6%2B10%7D%7B2%7D%2C%5C%3A%5Cfrac%7B9%2B5%7D%7B2%7D%5Cright%29)
![\left(x,\:y\right)=\left(8,\:7\right)](https://tex.z-dn.net/?f=%5Cleft%28x%2C%5C%3Ay%5Cright%29%3D%5Cleft%288%2C%5C%3A7%5Cright%29)
Therefore, the mid-point between the endpoints (10,5) and (6,9) is: