Solving Trig Functions: How Do You Solve For Sin 7.5 Using Half-Angle and Double-Angle Identities?

Formulas to remember:

sin(2a)=2sin(a)cos(a)
cos(2a)=cos²(a)-sin²(a)
From the second equation, replace a by a/2 and 2a by a to get
cos(a)=cos²(a/2)-sin²(a/2)
=2cos²(a/2)-1 ...(1)
=1-2sin²(a/2) ...(2)

From (1) we get
cos²(a/2)=(1/2)(1+cos(a))
From (2) we get
sin²(a/2)=(1/2)(1-cos(a))

So
sin(7.5)
=√((1/2)(1-cos(15))
=√((1/2)(1-√((1/2)(1+cos(30))))
=√((1/2)(1-√((1/2)(1+√(3)/2)))
=0.13...

dersagt

To solve for sin 7.5 using half-angle and double-angle identities, we need to use a combination of these trigonometric identities to simplify the expression.

1. Half-angle identity: sin(θ/2) = ±√[(1 - cosθ)/2]

2. Double-angle identity: sin(2θ) = 2sinθcosθ

Since we are given sin 7.5 and want to find an angle whose sine value is sin 7.5, we can rewrite sin 7.5 as sin (15/2) by using the angle-sum identity.

Now, let's solve for sin (15/2) using the half-angle identity:

sin (15/2) = ±√[(1 - cos 15)/2]

But to use this identity, we need to find cos 15.

To find cos 15, we can use the double-angle identity:

cos (2θ) = cos^2θ - sin^2θ

cos (30) = cos^2(15) - sin^2(15)

But we know that cos (30) = √3/2 and sin (30) = 1/2, so we can rewrite it as:

√3/2 = cos^2(15) - (sin 15)^2

Now, let's substitute cos^2(15) = √3/2 - (sin 15)^2 back into the half-angle identity for sin (15/2):

sin (15/2) = ±√[(1 - (√3/2 - (sin 15)^2))/2]

Now we have an equation in terms of sin (15/2), sin 15, and other constants.

To simplify further, we need to find the value of sin 15 in degrees. Since 15 is not a special angle, we'll need to use a calculator or a table that provides the value of sin 15.

Using a scientific calculator or trigonometric table, we find that sin 15 = 0.258819.

Plugging in this value and simplifying, we get:

sin (15/2) = ±√[(1 - (√3/2 - 0.258819^2))/2]

You can evaluate the expression further to get the final result.

Remember to check the quadrant of the angle (15/2) to determine the sign, which will be positive or negative.