Trigonometric Ratios Of Some Specific Angles

Trigonometric ratios of some common angles such as 0°, 45°, 90°, etc. are used very often in solving trigonometric questions.

Example 1: Find the value of cos 2A if A = 30°. Solution:

It is given that A = 30°

∴ cos 2A = cos (2 × 30°)

= cos 60°

=https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/179/455/951/10.1.8.2.1_ok_SS_html_eeecab0.gif

Example 2: Prove that sin 2A = 2 sin A cos A, for A = 45°.

Solution:

L.H.S = sin 2 A = sin (2 × 45°)

= sin 90 °

= 1

R.H.S. = 2 sin A cos A = 2 × sin 45° × cos 45°

= 2 ×https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/179/455/951/10.1.8.2.1_ok_SS_html_m1ed6a80.gif

= 1

L.H.S = R.H.S.

∴ sin 2A = 2 sin A cos A

Example 3: From the given figure XYZ, find ∠YZX and ∠ZXY.

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/179/455/951/10.1.8.2.1_ok_SS_html_1ff08916.png

Solution:

It is given that XY =https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/179/455/951/10.1.8.2.1_ok_SS_html_m632093a2.gif and YZ = 3

Now,https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/179/455/951/10.1.8.2.1_ok_SS_html_m6ba0e165.gif

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/179/455/951/10.1.8.2.1_ok_SS_html_17778754.gif

∴ ∠YZX = 30°

By angle sum property in ΔXYZ, we obtain

∠ZXY = 180° − 30° − 90°

∴ ∠ZXY = 60°

Example 4: If sin (A + B) = 1 and tan (A −B) = 0, where 0°< A + B ≤ 90°, then find A and B.

Solution:

sin (A + B) = 1,

We know that sin 90° = 1

⇒ A + B = 90° ... (1) tan (A −B) = 0

And, we know that tan 0° = 0

⇒ A − B = 0° ... (2)

On adding (1) and (2), we obtain 2A = 90°

⇒ A = 45°

On putting this value of A in (1), we obtain B = 45°

Thus, the value of both A and B is 45°.

Example 5: Find the value of https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/179/455/951/10.1.8.2.1_ok_SS_html_m2b8e0511.gif, if A = 60° and B = 30°.

Solution:

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/179/455/951/10.1.8.2.1_ok_SS_html_m342dcb0d.gif=https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/179/455/951/10.1.8.2.1_ok_SS_html_3c737560.gif

=https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/179/455/951/10.1.8.2.1_ok_SS_html_2f997f63.gif

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/179/455/951/10.1.8.2.1_ok_SS_html_77dac07.gif

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/179/455/951/10.1.8.2.1_ok_SS_html_13245ac4.gif

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/179/455/951/10.1.8.2.1_ok_SS_html_m139f53e4.gif

Example 6: Find the value of sin2 60° + 2 cos2 30° – tan2 45° + sec2 30°

Solution:

sin2 60° + 2 cos2 30° – tan2 45° + sec2 30°

https://img-nm.mnimgs.com/img/study_content/editlive_lp/56/2013_05_10_16_54_48/mathmlequation2475658389767669918.png

https://img-nm.mnimgs.com/img/study_content/editlive_lp/56/2013_05_10_16_54_48/mathmlequation1670015166518127301.png

Example 7: If sin x = sin 60° cos 30° + cos 60° sin 30° then find the value of x.

Solution:

We have

sin x = sin 60° cos 30° + cos 60° sin 30°

https://img-nm.mnimgs.com/img/study_content/editlive_lp/56/2013_05_10_16_54_48/mathmlequation4715126944157575725.png

⇒ sin x = 1

⇒ sin x = sin 90°

x = 90°

Example 8: ABCD is a trapezium such that AB || CD and AD = BC.

https://img-nm.mnimgs.com/img/study_content/editlive_lp/56/2013_05_10_16_54_48/12.png

Find the area of the trapezium.

Solution:

In right-angled ΔAED, we have

https://img-nm.mnimgs.com/img/study_content/editlive_lp/56/2013_05_10_16_54_48/mathmlequation2604357988298393806.png

In ΔAED and ΔBFC, we have

AD = BC

∠AED = ∠BEC = 90° and

AE = BF (Perpendiculars drawn between two parallel lines)

∴ΔAED https://img-nm.mnimgs.com/img/study_content/editlive_lp/56/2013_05_10_16_54_48/mathmlequation6659279449193421335.png ΔBFC

By CPCT, we have

ED = FC = 83√83cm

ABFE is a rectangle, so we have AB = FE = 12 cm

Now, DC = ED + FE + FC = https://img-nm.mnimgs.com/img/study_content/editlive_lp/56/2013_05_10_16_54_48/mathmlequation1320099958932559122.png + 12 cm + https://img-nm.mnimgs.com/img/study_content/editlive_lp/56/2013_05_10_16_54_48/mathmlequation1320099958932559122.png = https://img-nm.mnimgs.com/img/study_content/editlive_lp/56/2013_05_10_16_54_48/mathmlequation6832598049325752856.png

Area of trapezium ABCD = (AB + DC) × AE

https://img-nm.mnimgs.com/img/study_content/editlive_lp/56/2013_05_10_16_54_48/mathmlequation3448069877196565637.png