Math Problem Statement

This question requires you to fill in the blanks. Some steps will refer to previous answers you have typed, so it is necessary to "submit" after filling in each blank. The quickest way to do this is hit enter after typing instead of pushing the submit button.

Please see formatting instructions below the problem.

Prove the identity
sin ( α + β ) + sin ( α − β )

2 sin ( α ) cos ( β )

We will work on the left side:	  

sin ( α + β ) + sin ( α − β )

=

  1. Use sum and difference identities (type "alpha" for
    α and "beta" for
    β ):

=

Now you have:	

2 sin ( α ) cos ( β )

=

  1. Simplify:

=

Proven! Now you have:	  
  	 

FORMATTING INSTRUCTIONS BELOW Type in lowercase and use parentheses. For example type sin(x) not sinx.

For fractions use the divide symbol "/". For example sin(x)/cos(x) for
sin ( x ) cos ( x )

For multiplication of fractions use parentheses. For example (cos(a)/sin(b))(tan(c)/sin(d)) for
( cos ( a ) sin ( b ) ) ( tan ( c ) sin ( d ) ) .

For multiplication without fractions no symbol is necessary. For example cos(a)sin(b)

For exponents use "^". Parentheses will be necessary depending on how you choose to type your answer For example sin^2(x) for
sin 2 ( x ) . Or (sin(x))^2 for
( sin ( x ) ) 2 . For fractions with multiple terms in the numerator or denominator parentheses are necessary. For example (1+sin(x))/cos(x) for
1 + sin ( x ) cos ( x ) . Or (1+sin(x))/(1-sin(x)) for
1 + sin ( x ) 1 − sin ( x )

Solution

We start by applying the sum and difference identities for sine:

  1. Use sum and difference identities:

sin(α+β)+sin(αβ)=sin(α)cos(β)+cos(α)sin(β)+sin(α)cos(β)cos(α)sin(β)\sin(\alpha + \beta) + \sin(\alpha - \beta) = \sin(\alpha)\cos(\beta) + \cos(\alpha)\sin(\beta) + \sin(\alpha)\cos(\beta) - \cos(\alpha)\sin(\beta)

  1. Now simplify:

=sin(α)cos(β)+sin(α)cos(β)=2sin(α)cos(β)= \sin(\alpha)\cos(\beta) + \sin(\alpha)\cos(\beta) = 2\sin(\alpha)\cos(\beta)

  1. Therefore, we have proven:

sin(α+β)+sin(αβ)=2sin(α)cos(β)\sin(\alpha + \beta) + \sin(\alpha - \beta) = 2\sin(\alpha)\cos(\beta)

The identity is now fully proven!


Do you want details or have any questions?
Here are 5 related questions:

  1. What is the cosine sum identity for cos(α+β)\cos(\alpha + \beta)?
  2. How do you derive the sine double-angle identity from this result?
  3. Can you prove the identity cos(α+β)+cos(αβ)\cos(\alpha + \beta) + \cos(\alpha - \beta)?
  4. What are some applications of sum and difference identities in physics?
  5. How do these identities help in simplifying trigonometric integrals?

Tip: Sum and difference identities are essential for simplifying complex trigonometric expressions and proving various trigonometric equations.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Sum and Difference Identities

Formulas

sin(α + β) = sin(α)cos(β) + cos(α)sin(β)
sin(α − β) = sin(α)cos(β) − cos(α)sin(β)
sin(α + β) + sin(α − β) = 2sin(α)cos(β)

Theorems

Sum and Difference Formulas for Sine

Suitable Grade Level

Grades 10-12