Tips for Solving Half Angle Identities with Horizontal Shifts

[dranseth]

Homework Statement

I'm on the last section of identities entitled half angle identities. This one seems to give me some trouble because I have never encountered one with a horizontal shift in it. Tips?

tan 1/2( ß + π/2 ) = ( 1 + sin ß ) / cos ß

 

[quantumdude]

Recall that:

$$\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$$

Now, what do the graphs of the sine and cosine functions look like when you shift them to the left by $\frac{\pi}{2}$ rad?

 

[Architeuthis Dux]

I would suggest breaking down the problem into smaller parts and using known identities to solve it. First, we can rewrite the given equation using the double angle identity for tangent:

tan 1/2( ß + π/2 ) = sin( ß + π/2 ) / cos( ß + π/2 )

Next, we can use the sum identities for sine and cosine to simplify the equation:

tan 1/2( ß + π/2 ) = ( sin ß cos π/2 + cos ß sin π/2 ) / ( cos ß cos π/2 - sin ß sin π/2 )

Since cos π/2 = 0 and sin π/2 = 1, the equation becomes:

tan 1/2( ß + π/2 ) = sin ß / cos ß = tan ß

Now, we can use the half angle identity for tangent to solve for the left side of the equation:

tan 1/2( ß + π/2 ) = ( 1 - cos ß ) / sin ß

Substituting this into the original equation, we get:

( 1 - cos ß ) / sin ß = ( 1 + sin ß ) / cos ß

This can be simplified to:

1 - cos ß = ( 1 + sin ß ) sin ß / cos ß

Now, we can use the Pythagorean identity to simplify the right side of the equation:

1 - cos ß = ( 1 + sin ß ) √(1 - cos^2 ß) / cos ß

Simplifying further, we get:

1 - cos ß = ( 1 + sin ß ) √(sin^2 ß) / cos ß

Finally, we can use the identity sin^2 ß = 1 - cos^2 ß to simplify the equation to:

1 - cos ß = ( 1 + sin ß ) ( 1 - cos^2 ß ) / cos ß

This simplifies to the original equation, proving its validity. It may seem like a long process, but breaking down the problem into smaller steps and using known identities can help solve even the most challenging problems.