4.5: Proving Trigonometric Identities - Period 3/4

Today we learned about proving Trigonometric Identities, which is an equation with two side balance, and is true for all angles.

How?

Transfer the expression on one or both sides so that it equals the same expression on the other side.

Remember the basic identities, Pythagorean, Quotient, and Reciprocal identities? These are the basic of the trig identities and will be used, make sure to remember them or keep your pink sheet handy.

Also we learned about Double Angle Formulas, which is to express the sine or cosine of a double angle by retelling to its original angel (eg. sin(42^o ) = angel 21^o .

We recall: sin(a+b)= sin(a)cos(b )+ cos(a)sin(b)

(we set a = b; it’s the same, just double)

sin (x+x) = sin (x)cos(x) + cos(x)sin(x)

sin(2x) = 2sin(x)cos(x)

We also recall: cos(a+b) = cos(a)cos(b) – sin(a)sin(b)

(we set a = b; same, just double)

cos(x+x) = cos(x)cos(x) – sin(x)sin(x)

cos (2x) = cos^2 (x) – sin^2 (x)

Using Pythagorean Identity with cos:

1) cos (2x) = (1 – sin² (x)) – sin^2 (x)

cos(2x) = 1 – 2sin² (x)

2) cos(2x) = cos² (x) – (1 – cos² (x))

cos(2x) = 2cos² (x) – 1

Ex. 1) Using Double Angle to rewrite each expression.

cos(10x)

= cos[2(5x)]

= 1 – 2sin²(5x) or 2cos²(5x) – 1 or cos²(5x) – sin²(5x)

Ex. 2) Express as a single sine/cosine function.

6sinθcosθ

=3·2sinθcosθ

=3sin(2θ)

Ex. 3) Prove the Identity.

cos(x+y)cos(x-y) = cos²x + cos²y – 1

LS:

=(cos(x)cos(y) – sin(x)sin(y)) (cos(x)cos(y) + sin(x)sin(y))

=cos² (x)cos²(y) – sin²(x)sin²(y)

=cos²(x)cos²(y) – (1 – cos²(x))sin²(y)

=cos²(x)cos²(y) – sin²(y) + sin²(y)cos²(x)

=cos²(x) (cos²(y) + sin²(y)) – sin²(y)

=cos²(x) – sin²(y)

=cos²(x) – (1 – cos²(y))

=cos²(x) – 1 + cos²(y)

=cos²x + cos²y – 1

The example about was just one of the few other examples done in class.

*Side Note: we learned that Q.E.D (quod erat demonstrandum) means "Which was to be proven/demonstrated” in Latin, which goes at the end of an answer of a question*