Trigonometric Identities 
Proving Trigonometric Identities:
Since we are asked to prove that the identity is true, we can't assume it is.
We pick a side, indicate it with LHS or RHS then show it's the same as the other side.
RHS and LHS stand for Right Hand Side and Left Hand Side.
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TRIG IDENTITIES
cos^{ 2} x + sin^{ 2} x = 1  sec^{ 2} x = 1 + tan^{ 2} x  csc^{ 2} x = 1 + cot^{ 2} x 
cos^{ 2} x = 1 – sin^{ 2} x  tan^{ 2} x = sec^{ 2} x – 1  cot^{ 2} x = csc^{ 2} x – 1 
sin^{ 2} x = 1 – cos^{ 2} x  
csc x = 1/sin x sin x = 1 / csc x 
sec x = 1/ cos x cos x = 1 / sec x 
cot x = 1/ tan x tan x = 1 / cot x 
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memory tweak: The reciprocal functions are easy to remember if you notice that
only one in each pair has the "co" syllable.
Sine and COsecant are reciprocals. Sine doesn't have the "CO" syllable  so reciprocal does.
COsine and Secant are reciprocals. Cosine has the "CO" syllable  so reciprocal doesn't.
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Hint1: With a mix of sine, cosine and tangent, change them to sines and/or cosines.
Hint2: With a complex or multilevel fraction, flip and multiply like step 3 below.
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Example 1
Prove that
L.H.S.
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Here's one where we multiply by a unit fraction to prove it.
This operation is similar to rationalizing a binomial denominator in that
we multiply by the conjugate expression. (same terms, opposite sign).
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Example 2
Prove that
We will multiply LHS by a unit fraction like this:
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Sometimes we must reduce a mess of trig functions to
an expression with a single function using basic trig identities.
Example 3
Reduce cos A ( 1 + tan² A) to a single trig function of angle A.
cos A ( 1 + tan^{ 2} A) = cos A ( sec^{ 2} A) = cos A (1 / cos^{ 2} A) = sec A.
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Example 4
Reduce to a single trig function of angle B
( basics, composites, halfangle, practice, solutions )
Composite Angle Identities:
Use these to find the trig function values for a sum, difference or multiple of angles.
sin (A + B) = sin A cos B + cos A sin B  cos (A + B) = cos A cos B – sin A sin B 
sin (A – B) = sin A cos B – cos A sin B  cos (A – B) = cos A cos B + sin A sin B 
sin 2A = 2 sin A cos A  cos 2A = cos^{ 2} A – sin^{ 2} A 
 
 
** Note the relation between the formulae for sums and those for double angles.
If we replace B with A in the sum formula, we get 2A as the argument (angle).
So, knowing that sin 2A = 2 sin A cos A, we can extrapolate that
sin (A + B) is a sum of 2 terms, both of which have sin( ) and cos ( ).
Obviously, the blank brackets are filled with A and B, so
sin (A + B) must equal sin A cos B + cos A sin B.
And since addition is commutative, the order doesn't matter.
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Look at cos 2A = cos^{ 2} A – sin^{ 2} A.
This means cos (A + A) = cos A cos A – sin A sin A.
Change one of the A's in each term to a B, and there's the formula for cos (A + B).
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**Note also the pattern of signs (plusses and minuses).
For sin and tan, the signs correspond.
Sin or tan of a sum (+) means there's a sum in the formula too.
The same is true for sin or tan of a difference (–).
Note that sin (A – B) = sin A cos B – cos A sin B with emphasis on the " – "
For cosine formulae, the signs are always opposite.
For cos of a sum (+), there's a difference (–) in the formula and vice versa.
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Example 5
Let's derive the halfangle formulae from the identity for Cos 2A
We'll solve cos 2A = 2 cos^{ 2} A – 1 for cos A. Note that angle A is half of angle 2A
cos 2A + 1 = 2 cos^{ 2} A which becomes ½(cos 2A + 1) = cos^{ 2} A .
taking square root of both sides, we get
In exactly the same manner, we can show that
From these, we derive that
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Example 6
Use a halfangle formula to show that cos 45º =
Since 45º = ½ ( 90 º ), we set A = 45º and 2A = 90º
We know from the unit circle that cos 90º = 0, so
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Example 7
Use the identities for sin (A + B) and sin (A – B) to find
a) sin 75º and b) sin 15º given that
since 75º = 30º + 45º, we'll use sin (30º + 45º) = sin 30º cos 45º+ cos 30º sin 45º
since 15º = 45º – 30º, we'll use sin (45º – 30º ) = sin 45º cos 30º– cos 45º sin 30º
( basics, composites, halfangle, practice, solutions )
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1) Find sin 2A, cos 2A, and tan 2A given cot A = 4 / 3 and A is in the 1st quadrant.
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2) Use a sum or difference formula and exact values (no calculators) to find.
(the exact values come from the 45°, 45°, 90° and 30°, 60°, 90° triangles)
State the formula you're using.
a) cos 75º  b) tan (105º)  c) cos 15º 
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3) Use a halfangle formula and exact values (no calculators) to find the following values.
(Indicate the formula you are using.)
a) cos 15º  b) tan 22.5º  c) sin 15º 
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4) Prove these identities
a) sin^{ 2} A(1 + cot^{ 2} A) = 1  b) sec A – tan A sin A = cos A  c) 
d)  e) 2 cosA – sin 2A cscA = 0  f) cos 2A + sin 2A tanA = 1 
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5) Find sin A/2, cos A/2, and tan A/2 given cot A = 4 / 3 and A is in the 3rd quadrant.
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6) Reduce to a single term.
a) cos^{ 2} 5A – sin^{ 2} 5A  b) 2 sin 3x cos 3x 
( basics, composites, halfangle, practice, solutions )
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1) Find sin 2A, cos 2A, and tan 2A given cot A = 4 / 3 and A is in the 1st quadrant.
(view composite angle formulae)
sin 2A = 2 sin A cos A = 2 (3/5)(4/5) = 24/25
cos 2A = cos^{ 2} A – sin^{ 2} A = 16/25 – 9/25 = 7/25
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2)
a) cos 75º = cos (30º + 45º) use cos (A + B) 
b) tan (105º) = tan (60º + 45º) use tan (A + B) 
c) cos 15º = cos (45º – 30º) use cos (A – B) 
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a) cos 15º = cos (30/2)º use halfangle cos formula 
b) tan 22.5º = tan (45/2)º use halfangle tan formula 
c) sin 15º = sin (30/2)º use halfangle sin formula 
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4)
a) sin^{ 2} A(1 + cot^{ 2} A) = 1 LHS sin^{ 2} A(csc^{ 2} A) = 1 = RHS 
b) sec A – tan A sin A = cos A LHS 
c)

d) 
e) 2 cosA – sin 2A cscA = 0

f) cos 2A + sin 2A tan A = 1 
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5) Find sin A/2, cos A/2, and tan A/2 given cot A = 4 / 3 and A is in the 3rd quadrant.
Using the halfangle formulae
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6) Reduce to a single term.
a) cos^{ 2} 5A – sin^{ 2} 5A = cos 2(5A) = cos 10A  b) 2 sin 3x cos 3x = sin 2(3x)= sin 6x 
( basics, composites, halfangle, practice, solutions )
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