h.melikyan/12001 double-angle and half-angle formulas dr.hayk melikyan departmen of mathematics and...
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H.Melikyan/1200 1
Double-Angle and Half-Angle Formulas
Dr .Hayk MelikyanDepartmen of Mathematics and CS
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sin2 = 2 sin cos
cos2 = cos2 – sin2 = 1 – 2sin2 = 2cos2 – 1
tan 2 = 2tan1
2tan
Double-Angle Identities
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Three Forms of the Double-Angle Formula
for cos2
2
2
22
sin212cos
1cos22cos
sincos2cos
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Power-Reducing Formulas
2cos1
2cos1tan
2
2cos1cos
2
2cos1sin
2
2
2
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Example
Write an equivalent expression for sin4x that does not contain powers of trigonometric functions greater than 1.
8
2cos33
8
2cos12cos42
42
2cos12cos21
4
2cos2cos21
2
2cos1
2
2cos1sinsinsin
2
224
xxx
xxxx
xxxxx
Solution
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sinx2 = ±
1 – cos x2
cosx2 = ±
1 + cos x2
tanx2 = ±
1 – cos x1 + cos x =
sin x1 + cos x =
1 – cos xsin x
where the sign is determined by the quadrant in which x2 lies.
Half-Angle Identities
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Find the exact value of cos 112.5°.
Solution Because 112.5° 225°/2, we use the halfangle formula for cos /2 with 225°. What sign should we use when we apply the formula? Because 112.5° lies in quadrant II, where only the sine and cosecant are positive, cos 112.5° < 0. Thus, we use the sign in the halfangle formula.
Text Example
cos112.5 cos225
2
1 cos225
2
1 22
2
2 2
4
2 2
2
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Half-Angle Formulas for:
cos1
sin
2tan
sin
cos1
2tan
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Verifying a Trigonometric Identity
Verify the identity: tan cot 2csc2 2
x xx
sin (1 cos ) sin (1 cos )2csc
(1 cos )(1 cos ) (1 cos )(1 cos )
x x x xx
x x x x
2
2sin2csc
1 cos
xx
x
sin sin2csc
1 cos 1 cos
x xx
x x
2
2sin2csc
sin
xx
x
12 2csc
sinx
x
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The Half-Angle Formulas for Tangent
1 cos
tan Quadrant I or III2 1 cos
1 costan Quadrant II or IV
2 1 cos
1 costan in any quadrant
2 sin
sintan in any quadrant
2 1 cos
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Example
2sin1)cos(sin 2
Verify the following identity:
2sin1cossin22
2
cossin22
2cos1
2
2cos1
coscossin2sin
)cos(sin22
2
Solution
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Product-to-Sum and Sum-to-Product Formulas
Product-to-Sum Formulas
)sin()sin(2
1sincos
)sin()sin(2
1cossin
)cos()cos(2
1coscos
)cos()cos(2
1sinsin
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Example
xx 2cos3cos
Solution
Express the following product as a sum or difference:
)5cos()cos(2
1
)23cos()23cos(2
1
2cos3cos
)cos()cos(2
1coscos
xx
xxxx
xx
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Express each of the following products as a sum or difference. a. sin 8x sin 3x b. sin 4x cos x
Solution The product-to-sum formula that we are using is shown in each of the voice balloons.
a.
sin 8x sin 3x 1/2[cos (8x 3x) cos(8x 3x)] 1/2(cos 5x cos 11x)
sin sin = 1/2 [cos( - ) - cos( + )]
sin cos = 1/2[sin( + ) + sin( - )] b.
sin 4x cos x 1/2[sin (4x x) sin(4x x)] 1/2(sin 5x sin 3x)
Text Example
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Evaluating the Product of a Trigonometric ExpressionDetermine the exact value of the expression 3
sin cos8 8
1sin sin
2 2 4
1 1
12 2
1 3 3sin sin
2 8 8 8 8
1 2 1
2 2
1sin cos sin( ) sin( )
2
2 1
2 2
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Sum-to-Product Formulas
2sin
2sin2coscos
2cos
2cos2coscos
2cos
2sin2sinsin
2cos
2sin2sinsin
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Example
xx 2sin4sin
Express the difference as a product:
xxxx
xxxxxx
3cossin22
6cos
2
2sin2
2
24cos
2
24sin22sin4sin
2cos
2sin2sinsin
Solution
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Example
xx 4sinsin Solution
Express the sum as a product:
2
3cos
2
5sin2
2
4cos
2
4sin24sinsin
2cos
2sin2sinsin
xx
xxxxxx
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Example
2cot
2tan
sinsin
sinsin yxyx
yx
yx
Verify the following identity:
Solution
2cot
2tan
2sin
2cos
2cos
2sin
2cos
2sin2
2cos
2sin2
sinsin
sinsin
yxyxyxyx
yxyx
yxyx
yxyx
yx
yx
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Example
0 0
Express the following as a product and if possible find
the exact value. cos75 cos 15
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Example
Verify that the following is an identity:
sin 3 sintan
cos3 cos
x xx
x x
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• To solve an equation containing a single trigonometric function:
• Isolate the function on one side of the equation.
sinx = a (-1 ≤ a ≤ 1 )
cosx = a (-1 ≤ a ≤ 1 )
tan x = a ( for any real a )
• Solve for the variable.
Equations Involving a Single Trigonometric Function
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y = cos x
x
y
1
–1
y = 0.5
–4 2–2 4
cos x = 0.5 has infinitely many solutions for – < x <
y = cos x
x
y
1
–1
0.5
2
cos x = 0.5 has two solutions for 0 < x < 2
Trigonometric Equations
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Solve the equation: 3 sin x 2 5 sin x 1.
Solution The equation contains a single trigonometric function, sin x.
Step 1 Isolate the function on one side of the equation. We can solve for sin x by collecting all terms with sin x on the left side, and all the constant terms on the right side.
3 sin x 2 5 sin x 1 This is the given equation.
3 sin x 5 sin x 2 5 sin x 5 sin x – 1 Subtract 5 sin x from both sides.
sin x -1/2
Divide both sides by 2 and solve for sin x.
2 sin x 1 Add 2 to both sides.
2 sin x 2 1 Simplify.
Text Example
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Solve the equation: 2 cos2 x cos x 1 0, 0 x 2.
The solutions in the interval [0, 2) are /3, , and 5/3.
Solution The given equation is in quadratic form 2t2 t 1 0 with t cos x. Let us attempt to solve the equation using factoring.
2 cos2 x cos x 1 0 This is the given equation.
(2 cos x 1)(cos x 1) 0 Factor. Notice that 2t2 + t – 1 factors as (t – 1)(2t + 1).
cos x 1/2
2 cos x 1 cos x 1 Solve for cos x.
2 cos x 1 0 or cos x 1 0
Set each factor equal to 0.
Text Example
x x 2 x
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Example
cos29cos7
Solve the following equation:
Solution:
n2
5,3,
1cos
9cos9
cos29cos7
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Example
Solve the equation on the interval [0,2)
Solution:
3
3
2tan
3
7
3
6
7
62
3
3
2tan
and
and
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Example
Solve the equation on the interval [0,2)
Solution:03cos2cos2 xx
0
0
1cos3cos
01cos03cos
0)1)(cos3(cos
03cos2cos2
x
xsolutionno
xx
xx
xx
xx
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Example
Solve the equation on the interval [0,2)
Solution:
3
5,
3
2
1cos
1cos2
sincossin2
sin2sin
x
x
x
xxx
xx
xx sin2sin