Flashcards · 16 cards

Trigonometric ratios

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Rule
sin, cos and tan in a right-angled triangle?
Answer
$\sin\theta = \dfrac{\text{opp}}{\text{hyp}}$ , $\cos\theta = \dfrac{\text{adj}}{\text{hyp}}$ , $\tan\theta = \dfrac{\text{opp}}{\text{adj}}$ . Name the sides from the angle $\theta$ .
Name the sides from θsin = opp ÷ hyp, cos = adj ÷ hyp, tan = opp ÷ adj
From the lesson: Sine, cosine and tangent
Know it
$\tan 40^\circ = \dfrac{12}{x}$ . How do you find $x$ ?
Answer
Multiply both sides by $x$ , then divide by $\tan 40^\circ$ : $x = \dfrac{12}{\tan 40^\circ}$ .
From the lesson: Sine, cosine and tangent
Know it
$\sin x = \frac35$ and $x$ is acute. How do you find $\cos x$ and $\tan x$ ?
Answer
Draw the triangle: opposite 3, hypotenuse 5, so the adjacent side is $\sqrt{25 - 9} = 4$ . Then $\cos x = \frac45$ and $\tan x = \frac34$ .
From the lesson: Sine, cosine and tangent
Know it
What are $\sec\theta$ , $\operatorname{cosec}\theta$ and $\cot\theta$ ?
Answer
$\sec\theta = \dfrac{1}{\cos\theta}$ , $\operatorname{cosec}\theta = \dfrac{1}{\sin\theta}$ and $\cot\theta = \dfrac{1}{\tan\theta}$ .
From the lesson: Sine, cosine and tangent
Rule
$\sin^2\theta + \cos^2\theta = \,?$
Answer
$1$ : Pythagoras in a triangle with hypotenuse 1. Rearranged: $\cos^2\theta = 1 - \sin^2\theta$ and $\sin^2\theta = 1 - \cos^2\theta$ .
The identityLegs cos θ and sin θ, hypotenuse 1: sin²θ + cos²θ = 1
From the lesson: Sine, cosine and tangent
Know it
$1 + \tan^2\theta = \,?$
Answer
$\sec^2\theta$ . It comes from dividing $\sin^2\theta + \cos^2\theta = 1$ by $\cos^2\theta$ .
From the lesson: Sine, cosine and tangent
Know it
$1 + \cot^2\theta = \,?$
Answer
$\operatorname{cosec}^2\theta$ . It comes from dividing $\sin^2\theta + \cos^2\theta = 1$ by $\sin^2\theta$ .
From the lesson: Sine, cosine and tangent
Know it
Write $\tan\theta$ in terms of $\sin\theta$ and $\cos\theta$ .
Answer
$\tan\theta = \dfrac{\sin\theta}{\cos\theta}$ , because $\dfrac{\text{opp}}{\text{hyp}} \div \dfrac{\text{adj}}{\text{hyp}} = \dfrac{\text{opp}}{\text{adj}}$ .
From the lesson: Sine, cosine and tangent
Know it
$\sin\theta = \frac{7}{25}$ and $\theta$ is acute. Find $\cos\theta$ using the identity.
Answer
$\cos^2\theta = 1 - \frac{49}{625} = \frac{576}{625}$ , so $\cos\theta = \frac{24}{25}$ : positive, because $\theta$ is acute.
From the lesson: Sine, cosine and tangent
Rule
Exact values of sin, cos and tan for $30^\circ$ and $60^\circ$ ?
Answer
$\sin 30^\circ = \frac12$ , $\cos 30^\circ = \frac{\sqrt3}{2}$ , $\tan 30^\circ = \frac{1}{\sqrt3}$ . $\sin 60^\circ = \frac{\sqrt3}{2}$ , $\cos 60^\circ = \frac12$ , $\tan 60^\circ = \sqrt3$ .
Two special trianglesHalf a square and half an equilateral triangle give every exact value
From the lesson: Exact values for 30°, 45° and 60°
Know it
Exact values of sin, cos and tan for $45^\circ$ ?
Answer
$\sin 45^\circ = \cos 45^\circ = \frac{\sqrt2}{2}$ (that is, $\frac{1}{\sqrt2}$ ), and $\tan 45^\circ = 1$ .
From the lesson: Exact values for 30°, 45° and 60°
Rule
$\sin\theta = \cos(\,?\,)$
Answer
$\sin\theta = \cos(90^\circ - \theta)$ , and $\cos\theta = \sin(90^\circ - \theta)$ .
Complementary anglesThe side opposite θ is adjacent to 90° − θ
From the lesson: Exact values for 30°, 45° and 60°
Rule
Which ratios are positive in each quadrant?
Answer
A S T C, anticlockwise from $0^\circ$ : All in the first, Sine in the second, Tan in the third, Cos in the fourth.
Which ratio is positiveA S T C, anticlockwise from 0°
From the lesson: Sine and cosine beyond 90°
Know it
What is $\sin 150^\circ$ ?
Answer
$\sin 30^\circ = \frac12$ . $150^\circ$ is in the second quadrant, where sine is positive.
From the lesson: Sine and cosine beyond 90°
Which method?
WAEC 2022 · Paper 1 · Q29
Given that $\sin(5x - 28)^\circ = \cos(3x - 50)^\circ$ , $0^\circ \le x \le 90^\circ$ , find the value of $x$ .
When is $\sin A = \cos B$ ?
Answer
When $A$ and $B$ are complementary: $A + B = 90^\circ$ . So $(5x - 28) + (3x - 50) = 90$ .
From the lesson: Exact values for 30°, 45° and 60°Try the question
Which method?
WAEC 2021 · Paper 1 · Q22
If $\tan\theta = \frac34$ and $180^\circ < \theta < 270^\circ$ , find the value of $\cos\theta$ .
A 3-4-5 triangle gives the size. What decides the sign?
Answer
The quadrant. $180^\circ$ to $270^\circ$ is the third, where only tan is positive, so $\cos\theta$ is negative.
From the lesson: Sine and cosine beyond 90°Try the question