Introduction to Trig Ratios and to Identity
Trig Ratios
\sin \theta = \dfrac{opp}{hyp}
\cos \theta = \dfrac{adj}{hyp}
\tan \theta = \dfrac{opp}{adj}
Reciprocal Identities
\csc \theta = \dfrac{1}{\sin \theta}
\sec \theta = \dfrac{1}{\cos \theta}
\cot \theta = \dfrac{1}{\tan \theta}
Pythagorean Identities
\sin^2 \theta + \cos^2 \theta = 1
Simplifying \cot x \sin x
ex Simplify cot \theta \times \sin \theta.
= \dfrac{\cos \theta}{\sin \theta} \times \sin \theta
= \cos \theta
Simplifying Trig Expression 1+\tan^2x
ex 1+\tan^2x
= 1+ (\dfrac{\sin \theta}{\cos \theta})^2
= 1 + \dfrac{\sin^2 \theta}{\cos^2 \theta}
= \dfrac{\cos^2 \theta + \sin^2 \theta}{\cos^2 \theta}
= \dfrac{1}{\cos^2 \theta}
= \sec^2 \theta
Simplifying Trig Expression \dfrac{1 + \sin \theta}{\cos^2 \theta}
ex Simplify \dfrac{1 + \sin \theta}{\cos^2 \theta}
= \dfrac{1 + \sin \theta}{1 - \sin^2 \theta}
= \dfrac{1 + \sin \theta}{(1 + \sin \theta)(1 - \sin \theta)}
= \dfrac{1}{1 - \sin \theta}
Simplifying Trig Expression
\displaystyle \begin{aligned} & \tan ^{2} \theta-\sec ^{2} \theta \=& \frac{\sin ^{2} \theta}{\cos ^{2} \theta}-\frac{1}{\cos ^{2} \theta} \=& \frac{\sin ^{2} \theta-1}{\cos ^{2} \theta} \=& \frac{\sin ^{2} \theta-\left(\sin ^{2} \theta+\cos ^{2} \theta\right)}{\cos ^{2} \theta} \=& \frac{\sin ^{2} \theta-\sin ^{2} \theta-\cos ^{2} \theta}{\cos ^{2} \theta}=-1 \end{aligned}
Introduction to Proving Identities
LS = RS
Prove that LS is the same as RS.
ex \dfrac{1}{\cos \theta} + \tan \theta = \dfrac{1 + \sin \theta}{\cos \theta}
\displaystyle \begin{aligned} L S &=\frac{1}{\cos \alpha}+\frac{\sin \alpha}{\cos \alpha} \\ &=\frac{1+\sin \alpha}{\cos \alpha}=R .5 . \end{aligned}
Proving Identity with Relative Angles
Proving Identity ex2
ex Prove 2 \cos^2 \theta + \sin^2 \theta - 1 = \cos^2 \theta
Proving Identity ex3
ex Prove \tan \theta + \dfrac{1}{\tan \theta} = \dfrac{1}{\sin \theta \cos \theta}
\displaystyle \begin{aligned} & \tan \theta+\frac{1}{\tan \theta}=\frac{1}{\sin \theta \cos \theta} \\ L S &=\frac{\sin \theta}{\cos \theta} s+\frac{\cos \theta}{\sin \theta} \frac{c}{c} \\ &=\frac{\sin ^{2} \theta+\cos ^{2} \theta}{\sin \theta \cos \theta} \\ &=\frac{1}{\sin \theta \cos \theta}=R S \end{aligned}
For the proof of \sin^2B + \cos^2B = 1 using the triangle:
Fill in the missing step to prove.
\displaystyle
\begin{array}{lllll}
& LS &= (\frac{y}{r})^2 + (\frac{x}{r})^2\\
& &= \frac{y^2}{r^2} + \frac{x^2}{r^2}\\
& &= \frac{y^2+ x^2}{r^2} \\
& &= ......... \\
& &=1
\end{array}
Graph the relation y = \sin^2x + \cos^2x .
Prove the identity:
\sin \theta = \cos \theta \tan \theta
Which of the following could be a step going from right to left?
Prove the identity:
\displaystyle
\csc \theta = \sec \theta \cot \theta
Prove the identity:
\displaystyle
\cos \theta = \sin \theta \cot \theta
Which of the following could be a step going from right side to left?
Prove the identity:
\displaystyle
\sec \theta = \csc \theta \tan \theta
Which of the following could be a step going from right side to left?
Prove the identity:
\displaystyle
1 +\csc \theta = \csc \theta(1 + \sin \theta)
Prove the identity:
\displaystyle
\cot \theta \sin \theta \sec \theta = 1
Which of the following could be a step going from left side to right?
Prove the identity:
\displaystyle
\cos \theta(\sec \theta - 1) = 1 -\cos \theta
Which of the following could be a step going from left side to right?
Prove the identity:
\displaystyle
1 + \sin \theta = \sin \theta(1 + \csc \theta)
Prove that
\displaystyle
1 - \sin^2\theta = \sin \theta \cos \theta \cot \theta
Which of the following could be a step going from right side to left?
Prove that
\displaystyle
\csc^2\theta = \cot^2\theta + 1
Which of the following could be a step going from left side to right?
Prove that \displaystyle \frac{\cos \theta}{1 + \sin \theta} = \frac{1 - \sin \theta}{\cos \theta}
Start the proof by adding the rational expression on the left for \displaystyle \frac{\cos \theta}{1 - \sin \theta} + \frac{\cos \theta}{1 + \sin \theta}= \frac{2}{\cos \theta}.
Show your step and simply.
Prove that \displaystyle \csc^2\theta \cos^2\theta = \csc^2\theta - 1.
Prove that \displaystyle \tan \theta +\cot \theta = \frac{\sec \theta}{\sin \theta}.
Which of the following could be a step going from left side to right?
Prove that \displaystyle \cot^2 \theta(1 +\tan^2\theta) =\csc^2\theta.
Which of the following could be a step going from left side to right?
You are on the last leg of the orienteering course. Determine the direction and distance from the information given. Draw the leg on your map. Label all angles and distances.
Direction: East of south
Determine the two angles between 0° and 360^o such that \frac{\csc \theta}{\sec \theta} = \cot \theta. Add their degree measures and divide by 9. Use this angle.
Hint Use identities to simplify each side of the equation first.
Distance: The result of evaluating
\displaystyle
20(\sec^2\theta \sin^2\theta + \sec^2\theta \cos^2\theta -\tan^2\theta\sin^2\theta - \tan^2\theta \cos^2\theta)
, founded to the nearest metre if necessary.
Draw a right angle. Using the vertex as the centre, draw a quarter-circle that intersects the two arms of the angle. Select any point A on the quarter-circle, other than a point on one of the arms. Draw a tangent line to the quarter-circle at A such that the line intersects one arm at point B and the other arm at point C. Label \angle AOC as 6. Show that the measure of BA divided by the radius of the quarter-circle equals the cotangent of \angle \theta.
A student writes the following proof for the identity \cos \theta = \sin \theta \cot \theta. Critique it for form, and rewrite it in proper form.
Consider the equation
\displaystyle \tan^2\theta -\sin^2\theta = \sin^2\theta\tan^2\theta.
a) Which of the basic identities will you use first to simplify the left side?
b) Which is the one step before simplifying to the right side?
If \sin^2\theta + \sin^22\theta+ \sin^23\theta = 1, what does \cos^23\theta + \cos^22\theta + \cos^2\theta equal?
Given \sin \theta\cot \theta = \frac{\sqrt{3}}{2}, find the value of \sin\theta .
Given \cos \theta = 3\sin \theta, a possible value for \cos^2\theta is
\displaystyle
\begin{array}{cccccc}
&(A)&0.25 &(B)& 0.7071 &(C) & 0.5 &(D) & 0.9 &(E) & 1 \\
\end{array}