We proved that formula using the trigonometry relations sin3A + sinA = 4sinAcos^2A.
In the given question,
We have to prove the formula sin 3A+sin A = 4sinAcos^2A
The given expression is sin 3A+sin A = 4sinAcos^2A
To prove the formula we take the left side terms to the right side terms
The left side is sin 3A+sin A.
As we know that sin 3A = 3sinA − 4sin^3A
To solve the left side we put the value of sin 3A in sin 3A+sin A.
=sin 3A+sin A
=3sinA − 4sin^3A+sin A
Simplifying
= (3sinA+sin A) − 4sin^3A
= 4sinA − 4sin^3A
Taking 4sinA common from both terms
= 4sinA(1 − sin^2A)
As we know that cos^2A=1 − sin^2A. So
= 4sinAcos^2A
We proved the right hand side.
To learn more about formula of trigonometry terms link is here
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First find slope
slope=m=(9-6)/(-5-3)
m=-3/8
now equation
(y-9)=-(3/8)*(x+5)
8y-72=-3x-15
3x+8y=57
i can try to help
Step-by-step explanation:
