Sines and Cosines: Adding and Subtracting Angles

 Sines and Cosines:   Adding and Subtracting Angles

Note:   

π/2 radians = 90°,   π radians = 180°

Sine

sin(x + π/2) = sin(x) cos(π/2) + cos(x) sin(π/2) = cos(x)

sin(x - π/2) = sin(x) cos(π/2) - cos(x) sin(π/2) = -cos(x)

sin(π/2 - x)  = sin(π/2) cos(x) - cos(π/2) sin(x) = cos(x)

sin(x + π) = sin(x) cos(π) + cos(x) sin(π) = -sin(x)

sin(x - π) = sin(x) cos(π) - cos(x) sin(π) = -sin(x)

sin(π - x)  = sin(π) cos(x) - cos(π) sin(x) = sin(x)

Cosine

cos(x + π/2) = cos(x) cos(π/2) - sin(x) sin(π/2) = -sin(x)

cos(x - π/2) = cos(x) cos(π/2) + sin(x) sin(π/2) = sin(x)

cos(π/2 - x) = cos(π/2) cos(x) + sin(π/2) sin(x) = sin(x)

cos(x + π) = cos(x) cos(π) - sin(x) sin(π) = -cos(x)

cos(x - π) = cos(x) cos(π) + sin(x) sin(π) = -cos(x)

cos(π - x) = cos(π) cos(x) + sin(π) sin(x) = -cos(x)

Eddie

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sin(atan x), sin(acos x), cos(asin x), cos(atan x), tan(asin x), tan(acos x)

Deriving trig functions where the angle is an inverse trig function. Example: sin(acos x).

Inverse Functions

sin(asin t) = t
cos(acos t) = t
tan(atan t) = t

Other such calculations can be derived from the three Pythagorean trig identities:

sin^2 x + cos^2 x = 1
sec^2 = 1 + tan^2 x
1 + cot^2 x = csc^2 x

sin^2 x + cos^2 x = 1

Also known as (sin x)^2 + (cos x)^2 = 1

Let x = asin t where -1 ≤ t ≤ 1
(sin(asin t))^2 + (cos(asin t))^2 = 1
t^2 + (cos(asin t))^2 = 1
(cos(asin t))^2 = 1 - t^2
cos(asin t) = √(1 - t^2)

Let x = acos t, -1 ≤ t ≤ 1
(sin(acos t))^2 + (cos(acos t))^2 = 1
(sin(acos t))^2 = 1 - (cos(acos t))^2
(sin(acos t))^2 = 1 - t^2
sin(acos t) = √(1 - t^2)

sec^2 x = 1 + tan^2 x

sec^2 x = 1 + tan^2 x
1/cos^2 x = 1 + tan^2 x

Let x = atan t, -1 ≤ t ≤ 1

1/(cos(atan t))^2 = 1 + (tan(atan t))^2
1/((cos(atan t))^2 = 1 + t^2
(cos(atan t))^2 = 1/(1 + t^2)
cos(atan t) = √(1/(1 + t^2))

Let x = acos t, -1 ≤ t ≤ 1

1/(cos(acos t))^2 = 1 + (tan(acos t))^2
1/t^2 = 1 + (tan(acos t))^2
(tan(acos t))^2 = 1/t^2 - 1
(tan(acos t))^2 = (1 - t^2)/t^2
(tan(acos t)) = √(1 - t^2)/t

1 + cot^2 x = csc^2 x

1 + cot^2 x = csc^2 x
1 + 1/tan^2 x = 1/sin^2 x

Let x = asin t, -1 ≤ t ≤ 1

1 + 1/(tan(asin t))^2 = 1/(sin(asin t))^2
1/(tan(asin t))^2 = 1/t^2 - 1
1/(tan(asin t))^2 = (1 - t^2)/t^2
(tan(asin t))^2 = t^2/(1 - t^2)
tan(asin t) = t/√(1 - t^2)

Let x = atan t, -1 ≤ t ≤ 1

1 + 1/(tan(atan t))^2 = 1/(sin(atan t))^2
1 + 1/t^2 = 1/(sin(atan t))^2
(t^2 + 1)/t^2 = 1/(sin(atan t))^2
(sin(atan t))^2 = t^2/(t^2 + 1)
sin(atan t) = t/√(t^2 +1)

Summary:

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cos(asin t) = √(1 - t^2)
sin(acos t) = √(1 - t^2)
cos(atan t) = √(1/(1 + t^2))
tan(acos t) = √(1 - t^2)/t
tan(asin t) = t/√(1 - t^2)
sin(atan t) = t/√(t^2 +1)

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This blog is property of Edward Shore. 2014

Why sin^2 x + cos^2 x = 1

Why does sin^2 x + cos^2 x = 1?  Find out in this video. 

 

Link:  http://youtu.be/0ger4IUFhPg

 

- Eddie

 

 

 

 

Calculus Revisited #2: Working with Trigonometric Functions

Welcome to Part 2 of a series of the 21 Part Series Calculus Revisited, where we cover some of the major topics. Today's and tomorrow's blog entries will be cover the various transcendental functions: today trigonometric functions, tomorrow exponential and logarithmic functions.

Trigonometric Functions

Trigonometric Functions include sine, cosine, and tangent. (abbreviated sin, cos, and tan, respectively) They have corresponding reciprocal functions, cosecant, secant, and cotangent (abbreviated csc, sec, and cot respectively).

The relationship between the trigonometric and their reciprocal functions:

csc x = 1 / sin x
sec x = 1 / cos x
cot x = 1 / tan x

The inverse functions of sine, cosine, and tangent are the arcsine, arccosine, and arctangent functions, respectively. (abbreviated asin or sin^-1, acos or cos^-1, atan or tan^-1 respectively)

sin (asin x) = x
asin (sin x) = x
cos (acos x) = x
acos (cos x) = x
tan (atan x) = x
atan (tan x) = x


The graphs of sin x, cos x, and tan x follow.

Remember: sin^2 x = (sin x)^2. sin^-1 x = asin x. This applies for all trigonometric functions.

More Common Trig Identities

tan x = sin x/cos x
cot x = cos x/tan x

sin^2 x + cos^2 x = (sin x)^2 + (cos x)^2 = 1
tan^2 x + 1 = sec^2 x
1 + cot^2 x = csc^2 x

sin(π - x) = sin x
cos(π - x) = -cos x

sin 2x = 2 sin x cos x
cos 2x = 1 - 2 sin^2 x

sin(a + b) = cos a sin b + sin a cos b
cos(a + b) = cos a cos b - sin a sin b

Problems

1. Let y(x) = a sin (bx), where a and b are any real numbers. What is the significance of a and b?

a is the amplitude of the sine function, where a is the maximum value of y(x) and -a is the minimum value of y(x).

b involved in the period of y(x). The period is π/b.

Note: The variables of a and b have the same significance in the function y(x) = a cos (bx).

2. Is sin x an even or odd function? What about cos x and tan x?

The function f(x) is an even function if for every x, f(-x) = f(x). The function f(x) is an odd function if for every x, f(-x) = -f(x).

Using the graphs above as a reference, sin(-x) = -sin x and cos(-x) = cos x. Note that tan x = sin x/cos x. Then tan(-x) = sin(-x)/cos(-x) = - sin x/cos x = - tan x.

Therefore, sin x is an odd function, cos x is an even function, and tax x is an odd function.

3. Solve cos^2 x - cos x = sin^2 x.

Recall sin^2 x + cos^2 x = 1
Then sin^2 x = 1 - cos^2 x

cos^2 x - cos x = sin^2 x
cos^2 x - cos x = 1 - cos^2 x
2 cos^2 x - cos x - 1 = 0
2 (cos x)^2 - (cos x) - 1 = 0
This is quadratic equation in terms of cos x. This implies that
(cos x - 1) * (2 cos x + 1) = 0
Hence:
cos x - 1 = 0
cos x = 1
x = acos 1 = 0
and
2 cos x + 1 = 0
2 cos x = -1
cos x = -1/2
x = acos (-1/2) = 2π/3

The solutions are x = 0, 2π/3

4. Solve 2 sin x = sin 2x

Recall sin 2x = 2 sin x cos x

2 sin x = sin 2x
2 sin x = 2 sin x cos x
Since sin x can be zero, don't divide by sin x. Instead
2 sin x - 2 sin x cos x = 0
Factor out 2 sin x...
(2 sin x)(1 - cos x) = 0
Then
2 sin x = 0
sin x = 0
x = asin 0 = 0
and
1 - cos x = 0
1 = cos x
x = acos 1 = 0

The solution is x = 0.

Next time we will work exponential and logarithmic functions. Until then, take care! Eddie


This blog is property of Edward Shore. © 2012