Precalculus by Richard Wright
Whoever loves money never has enough; whoever loves wealth is never satisfied with their income. This too is meaningless. Ecclesiastes 5:10 NIV
Summary: In this section, you will:
SDA NAD Content Standards (2018): PC.5.1
The length of a shadow can be calculated from a complex trigonometric function based on the angle of elevation of the sun. Trigonometric identities can be used to simplify the function.
To verify trigonometric identities, use the fundamental identities to simplify one side of the identity to make it look like the other side of the equation.
Verify \(\left(1 - \tan α \right)\left(1 + \tan α \right) = 2 - \sec^2 α \).
The left side is more complicated, so start with that side.
$$\left(1 - \tan α \right)\left(1 + \tan α \right) = 2 - \sec^2 α $$
Multiply
$$\textcolor{blue}{1 - \tan^2 α} = 2 - \sec^2 α $$
Pythagorean identity (\(\tan^2 u + 1 = \sec^2 u\)) solved for tangent \(\tan^2 u = \sec^2 u - 1\).
$$1 - \left(\textcolor{red}{\sec^2 α - 1}\right) = 2 - \sec^2 α $$
Simplify
$$\textcolor{green}{2 - \sec^2 α} = 2 - \sec^2 α $$
Graph both sides of the identity on the same coordinate plane. The graphs will be identical if it is an identity.
Verify \(\sin^2 x - \sin^4 x = \cos^2 x - \cos^4 x\)
Both sides are equally complicated, so either side will be a good starting point. Let’s start with the right side.
$$\sin^2 x - \sin^4 x = \cos^2 x - \cos^4 x$$
Factor out cos2 x
$$\sin^2 x - \sin^4 x = \textcolor{blue}{\cos^2 x}\left(1 - \cos^2 x\right)$$
Use the Pythagorean identity (\(\sin^2 u + \cos^2 u = 1\)) solved for cos2 u (\(\cos^2 u = 1 - \sin^2 u\)).
$$\sin^2 x - \sin^4 x = \left(\textcolor{red}{1 - \sin^2 x}\right)\left(1 - \left(\textcolor{red}{1 - \sin^2 x}\right)\right)$$
Simplify
\(\sin^2 x - \sin^4 x = \left(1 - \sin^2 x\right)\left(\textcolor{green}{\sin^2 x}\right)\)
Multiply
$$\sin^2 x - \sin^4 x = \sin^2 x - \sin^4 x$$
Verify \(\tan^4 x - \sec^4 x = -1 - 2\tan^2 x\).
Answers will very.
Verify \(\frac{\tan^2 x}{\sec x} = \sec x - \cos x\).
Left side will be easier to work with.
$$\frac{\tan^2 x}{\sec x} = \sec x - \cos x$$
Use the Pythagorean identity (\(1 + \tan^2 u = \sec^2 u\)) solved for tan2 u (\(\tan^2 u = \sec^2 u - 1\)).
$$\frac{\textcolor{blue}{\sec^2 x - 1}}{\sec x} = \sec x - \cos x$$
Separate the fraction into two fractions.
$$\frac{\textcolor{red}{\sec^2 x}}{\sec x} - \frac{\textcolor{red}{1}}{\sec x} = \sec x - \cos x$$
Simplify the fractions and \(\frac{1}{\sec x} = \cos x\).
$$\sec x - \textcolor{green}{\cos x} = \sec x - \cos x$$
Verify \(\frac{1}{\csc x\cot x} = \sec x - \cos x\).
The left is slightly more complicated, so start there.
$$\frac{1}{\csc x\cot x} = \sec x - \cos x$$
Since the fraction is \(\frac{1}{something}\) it looks like a reciprocal. Use the reciprocal identities.
$$\textcolor{blue}{\sin x \tan x} = \sec x - \cos x$$
An identity that relates tangent and sine is \(\tan u = \frac{\sin u}{\cos u}\).
$$\sin x \textcolor{red}{\frac{\sin x}{\cos x}} = \sec x - \cos x$$
$$\frac{\textcolor{red}{\sin^2 x}}{\cos x} = \sec x - \cos x$$
Use the Pythagorean identity \(\sin^2 u + \cos^2 u = 1\) to substitute for sin2 x.
$$\frac{\textcolor{green}{1 - \cos^2 x}}{\cos x} = \sec x - \cos x$$
Separate the fraction into two fractions.
$$\frac{\textcolor{brown}{1}}{\cos x} - \frac{\textcolor{brown}{\cos^2 x}}{\cos x} = \sec x - \cos x$$
Simplify the fractions and \(\frac{1}{\cos x} = \sec x\).
$$\textcolor{purple}{\sec x} - \cos x = \sec x - \cos x$$
Verify \(\frac{1}{\cos x \tan x} = \csc x\)
Answers will very.
Verify \(\sin\left(-θ \right)\sec\left(-θ \right) = -\tan θ \).
The left side will be easier to work with.
$$\sin\left(-θ \right)\sec\left(-θ \right) = -\tan θ $$
Use the even/odd identities (\(\sin\left(-u\right) = -\sin u\) and \(\sec\left(-u\right) = \sec u\)).
$$\textcolor{blue}{-\sin θ \sec θ} = -\tan θ$$
Use the reciprocal identity (\(\sec u = \frac{1}{\cos u}\)) to rewrite sec θ.
$$-\sin θ \textcolor{red}{\frac{1}{\cos θ}} = -\tan θ$$
$$-\frac{\sin θ}{\cos θ} = -\tan θ$$
Use the quotient identity (\(\tan u = \frac{\sin u}{\cos u}\)).
$$\textcolor{green}{-\tan θ} = -\tan θ$$
Verify \(\frac{\sin x + \tan y}{\sin x \tan y} = \cot y + \csc x\).
Again the left side is more complex.
$$\frac{\sin x + \tan y}{\sin x \tan y} = \cot y + \csc x$$
Separate the fraction into two fractions.
$$\frac{\textcolor{blue}{\sin x}}{\sin x \tan y} + \frac{\textcolor{blue}{\tan y}}{\sin x \tan y} = \cot y + \csc x$$
Simplify.
$$\textcolor{red}{\frac{1}{\tan y}} + \textcolor{red}{\frac{1}{\sin x}} = \cot y + \csc x$$
Use reciprocal identities (\(\cot u = \frac{1}{\tan u}\) and \(\csc u = \frac{1}{\sin u}\)).
$$\textcolor{green}{\cot y} + \textcolor{green}{\csc x} = \cot y + \csc x$$
Verify \(\tan\left(\frac{π}{2} - x\right)\sin \left(-x\right) = -\cos x\).
Answers will very.
Verify \(1 - \sin^2\left(\frac{π}{2} - x\right) = \sin^2 x\) a) algebraically and b) graphically.
The left side is again the most complex.
$$1 - \sin^2\left(\frac{π}{2} - x\right) = \sin^2 x$$
Use the cofunction identity (\(\sin\left(\frac{π}{2} - u\right) = \cos u\)).
$$1 - \textcolor{blue}{\cos^2 x} = \sin^2 x$$
Use the Pythagorean identity (\(\sin^2 u + \cos^2 u = 1\)) solve for sin2 u (\(\sin^2 u = 1 - \cos^2 u\)).
$$\textcolor{green}{\sin^2 x} = \sin^2 x$$
Graph the left side of the equation, then graph the right side of the equation and notice that they are the same.
Verify \(\frac{\sec x - \csc x}{\sec x \csc x} = \sin x - \cos x\) both (a) algebraically and (b) graphically.
Answers will very;
Helpful videos about this lesson.
Verify the identities.
Verify the identity algebraically and graphically.
The length, ℓ, of a shadow cast by a vertical stick of height, h, when the angle of elevation of the sun is θ can be modeled by
$$ℓ = \frac{h \cos θ}{\cos\left(90° - θ\right)}$$
Verify that \(ℓ = h \cot θ\).
Mixed Review