bit rusty on these types of q can any1 explain how to approach them (1 Viewer)

[lmao1010]

 

[ExtremelyBoredUser]

View attachment 34945
View attachment 34946

latex broke lol

sin^-1(x) + \sin^-1(3x) = 0

sin(\sin^-1(x) + \sin^-1(3x)) = sin(0)

Let A,B be some dummy variable.

A = sin^-1(x) hence sin(A) = x/1 (use this for later)
B = sin^-1(3x) hence sin(B) = 3x/1 (use this for later)

Now simplifying the equation.

sin(A+B) = 0

sinAcosB + cosAsinB = 0 which is the equation we need to solve

x/1 = sin(A), construct a triangle

Opposite/Hyp = x/1

x is opposite side, 1 is hypotenuse



Find the other side by the pythagorean identity.

cos(A) = \sqrt{1-x^2} as cos is adj/hyp

Do this similar triangle method for B = sin^-1(3x) to find cos(B) and sin(B) and solve the trig equation above.