CBSE Class 10 Mathematics Introduction to Trigonometry Notes Set 17

Trigonometric Ratios of an Acute Angle

In a right triangle, the side opposite the angle θ is the perpendicular (P), the side next to it is the base (B) and the side opposite the right angle is the hypotenuse (H). The ratios are \( \sin\theta=\frac PH \), \( \cos\theta=\frac BH \), \( \tan\theta=\frac PB \), and their reciprocals \( \csc\theta=\frac HP \), \( \sec\theta=\frac HB \), \( \cot\theta=\frac BP \).
Also \( \tan\theta=\frac{\sin\theta}{\cos\theta} \) and \( \cot\theta=\frac{\cos\theta}{\sin\theta} \). If one ratio is given, draw the triangle, find the third side with the Pythagoras theorem and read off every other ratio.

Values for Standard Angles

\( \sin0^\circ=0,\ \sin30^\circ=\frac12,\ \sin45^\circ=\frac1{\sqrt2},\ \sin60^\circ=\frac{\sqrt3}2,\ \sin90^\circ=1 \).
Cosine takes the same values in the reverse order. \( \tan0^\circ=0,\ \tan30^\circ=\frac1{\sqrt3},\ \tan45^\circ=1,\ \tan60^\circ=\sqrt3 \), and tan 90° is not defined.

Complementary Angles

For an acute angle, \( \sin(90^\circ-\theta)=\cos\theta \), \( \cos(90^\circ-\theta)=\sin\theta \), \( \tan(90^\circ-\theta)=\cot\theta \), \( \sec(90^\circ-\theta)=\csc\theta \), \( \csc(90^\circ-\theta)=\sec\theta \) and \( \cot(90^\circ-\theta)=\tan\theta \).
These relations let you rewrite a ratio of a large angle as the co-ratio of a small angle, for example \( \sin69^\circ=\cos21^\circ \). A product such as \( \tan20^\circ\tan70^\circ \) equals 1 because the angles add to 90°.

Trigonometric Identities

\( \sin^2\theta+\cos^2\theta=1 \), \( 1+\tan^2\theta=\sec^2\theta \) and \( 1+\cot^2\theta=\csc^2\theta \). To prove an identity, work on the more complicated side, express everything in sine and cosine and simplify to the other side.
A useful trick for expressions like \( \sqrt{\frac{1+\sin A}{1-\sin A}} \) is to multiply the numerator and denominator by \( 1+\sin A \), which gives \( \frac{1+\sin A}{\cos A}=\sec A+\tan A \) for an acute angle.

Angles in Radians and Addition Formulas

\( \pi \) radians equal 180°, so degrees are converted to radians by multiplying by \( \frac{\pi}{180} \) and radians to degrees by multiplying by \( \frac{180}{\pi} \).
\( \sin(A\pm B)=\sin A\cos B\pm\cos A\sin B \) and \( \tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B} \) give values such as \( \sin15^\circ=\frac{\sqrt3-1}{2\sqrt2} \) and \( \tan75^\circ=2+\sqrt3 \).

Worked Examples

Worked Example 1. If \( \tan\theta=\frac{5}{12} \) (θ acute), find \( \sin\theta+\cos\theta \).
Answer:
Hypotenuse \( =\sqrt{25+144}=13 \), so \( \sin\theta=\frac5{13} \) and \( \cos\theta=\frac{12}{13} \).
\( \sin\theta+\cos\theta=\frac{17}{13} \).

Worked Example 2. Evaluate \( \frac{\sin35^\circ}{\cos55^\circ}+\tan10^\circ\tan80^\circ \).
Answer:
\( \cos55^\circ=\sin35^\circ \), so the first term is 1. \( \tan80^\circ=\cot10^\circ \), so the product is 1.
The value is 2.

Worked Example 3. Prove that \( (1+\tan^2A)\cos^2A=1 \).
Answer:
\( 1+\tan^2A=\sec^2A \).
\( \sec^2A\cos^2A=1 \), which is the right-hand side.