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Trigonometry Trigonometric Table for Standard Angles, Exercises of Mathematics

Mathematics

A series of trigonometric problems and their solutions involving standard angles. It includes the use of trigonometric functions such as cotangent, secant, sine, cosine, and tangent. The solutions involve the use of trigonometric identities and the trigonometric table for standard angles. The document also includes the verification of trigonometric equations and the evaluation of expressions involving trigonometric functions.

Typology: Exercises

Pre 2010

Available from 03/05/2022

SKarthigeyan
SKarthigeyan 🇮🇳

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Trigonometry
Trigonometric table for standard angles
1. Find the value of each of the following expressions.
(i) 4 cot 45 – sec2 60 + sin2 60 + cos2 90
Solution:
4 cot 45 – sec2 60 + sin2 60 + cos2 90
2
= 4 (1) – (2)2 + + 0
= 4 – 4 + =
(ii) sin 60 cos 30 – cos 60 sin 30 + 3cos 0 + 5 sin 90
Solution: sin 60 cos 30 – cos 60 sin 30 + 3cos 0 + 5 sin 90
= . – . + 3 (1) + 5 (1)
= – + 3 + 5
= + 8 = =
tan260 + 4 sin245 + sec230 + 5cos290
cosec(30) + sec(60) – cot2(30)
2 2
3 + 4 + + 5(0)2
Solution: =
2 + 2 – 3
3 + 2 +
= = 5 + = =
2 1 2
2 3
3 1
4 4
3
2
3 3
4 4
3 3 1 1
2 2 2 2
1 1 +16 17
2 2 2
(iii)
2
4 15 + 4 19
4 – 3 3 3 3
pf3
pf4
pf5

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Trigonometry

Trigonometric table for standard angles

1. Find the value of each of the following expressions. (i) 4 cot 45– sec^2 60  + sin^2 60  + cos^2 90  Solution: 4 cot 45 – sec^2 60  + sin^2 60  + cos^2 90  2 = 4 (1) – (2) 2

    • 0 = 4 – 4 + = (ii) sin 60cos 30– cos 60sin 30+ 3cos 0+ 5 sin 90Solution: sin 60 cos 30 – cos 60 sin 30 + 3cos 0 + 5 sin 90

tan^260  + 4 sin^245  + sec^230  + 5cos^290  cosec(30) + sec(60) – cot^2 (30) 2 2  3 + 4 + + 5(0)^2 Solution: = 2 + 2 –  3 3 + 2 + = = 5 + = =

(iii) 2 4 3 4 15 + 4 19 4 – 3 3 3 3

2. Verify that (i) sin 90= sin 60cos 30+ cos 60sin 30

Solution: 1 =. +.

(ii) cos 0= cos 2 45+ sin 2 45Solution: cos 0 = cos 2 45  + sin 2 45  1 = + = + = 1 Verified

3. If= 30verify that (i) sin 3= 3 sin– 4 sin^3  Solution: sin 3 = 3 sin  – 4 sin^3  L.H.S = sin 3 (30) = sin 90 = 1 R.H.S = 3 sin 30 – 4 sin 3 30  = 3 – 4 = – = – = = 1 Verified. (ii) cos 3= 4 cos^3  – 3 cosSolution: L.H.S = cos 3  = cos 3 (30) = cos 90 = 0 R.H.S = 4 cos 3  – 3 cos  = 4 cos 3 30  – 3 cos 30 = 4 – 3 = 4 – = – = 0

3 3 4 2 2 2 8 3 1 2 2 2 2  3  3 2 2 3 3  3 3  3 3  3 3  3

5. Solve the following equation for 0<   90, = 2 Solution: = 2 cos 2  – 3 cos  + 2 = 2 sin 2  cos 2  – 3 cos  + 2 – 2 sin 2  = 0 cos 2  – 3 cos + 2 (1 – sin 2 ) = 0 cos 2  – 3 cos  + 2 cos 2  = 0 3 cos 2  – 3 cos  = 0 3 cos (cos – 1) = 0 3 cos  = 0 ; cos  – 1 = 0 cos  = 0 ; cos  = 1 = 90 ;  =  (not possible). 6. When 0     90, solve the following: (i) 2 cos 3= 1 Solution: 2 cos 3  = 1 cos 3  = = cos 60

(ii) 2 cos 2+ sin– 2 = 0 Solution: 2 cos 2  + sin  – 2 = 0 2(1 – sin^2  + sin  – 2 = 0 2 – 2 sin 2  + sin  – 2 = 0 2 sin 2 sin  = 0 sin  (2 sin – 1) = 0 sin  = 0 2 sin  – 1 = 0 = sin sin  = = sin 30

cos 2  3cos+ 2 sin 2  cos 2  3cos + 2 sin^2  1 2 60  3 1 2

7. (i) If 4 tan= 5, evaluate Solution: 4 tan  = 5

tan  =

  by cos  in the Nr & Dr.

2 – 3 tan  3 + 2 tan  = = (ii) If cot= , find the value of. Solution: cot  = =

 by sin  in the Nr & Dr.

8. Find x from the following equation: tan^2 45  – cos^2 60  = x****. sin 45cos 45cot 30. 2

Solution: 1 – = x  3

 3  3 x  3 4 2 = x 2cos– 3sin3cos+ 2sin  5 4 2 cos – 3 sin 3 cos + 2sin = =

a cos– sinb cos+ sin  a b cos – sin cos + sin cot  – 1  cot  + 1 a b a – b a a + b b

1 x  3 4 2 =  3 3 x  3 4 2