Shafrina IrzaX.A

sin o

ah

a

cos a

ah

a

tan o

aa

a

oppo

site

oppo

site

oppo

site

adjacent

adjacent

adjacent

hypotenuse

hypotenuse

hypotenuse

sin o

bh

cos a

bh

tan o

ba

hypotenuse

hypotenuse

hypotenuse

adja

cent

adja

cent

adja

cent

opposite

opposite

opposite

b

b

b

1. Find the sine of 30 ° and the length of the adjacent side!

30 °

1) From your calculator or the trig tables, find that sin 30 ° = 0.5

2) Remember that sine = opposite/hypotenuse.3) Substitute 0.5 = opposite/184) Solve: opposite = 9

18

adja

cent

opposite

2. Find the cosine of a ° and the length of the opposite side!

a ° 1) Remember that cosine = adjacent/hypotenuse2) Cos a ° = 20/25 =0.83) From your calculator or the trig tables, find where the

cosine = .8, about 36.87 °.4) Angle a is 36.87°.5) The opposite side can be found using the Pythagorean

Theorem or another trig function.6) Sin a ° = opposite/hypotenuse 7) Sin 36.87° = opposite/25 (Trig tables or calculator)8) 0.6 = opposite/259) Opposite = 15

10)When using a calculator, the process is easier.11)Enter 20/25 =12)Answer is 0.813)Enter SHIFT KEY, COS-1 KEY, and = to get 36.87°. The

INVERSE KEYS of SIN-1 , COS-1 , TAN-1 switch the view from the trig table value to the degree value. Remember to use the SHIFT KEY or 2nd KEY to use inverse functions.

25

20

opposite

Most problems involving trip will have to do with finding the height of a ladder, a tree, or something leaning against something else at an angle, or finding how far away from an object someone is standing. These all use the standard trig functions of sin, cos, and tan to solve.

You’re standing 50 feet away from a tree, and using a clinometer, find that the top of the tree is at a 40° angle from where you stand. How high is the tree?

1)You’re looking for the opposite side, and you know the length of the adjacent side. Use the tangent.2)Tan 40° = opposite/adjacent3).8391 = opposite/50 (Multiply both sides by 50)4)Height of the tree is 42.95 feet

50

40°

00 300 450 600 900

sin θ 0 1 / 2 1 / √2 √3 / 2 1

cos θ 1 √3 / 2 1 / √2 1 / 2 0

tan θ 0 1 / √3 1 √3 ∞

cosec θ ∞ 2 √2 2 / √3 1

sec θ 1 2 / √3 √2 √2 ∞

cot θ ∞ √3 1 1 / √3 0

Applications of Trigonometrical Tables :(1) It is used in the measurement of the height of the flying aeroplanes.(2) It can be used for the measurement of depth of the sea water.(3) It can be used for measuring the length of the ladder.

Complementary angles.Two angles are complementary if they add up to 90 degrees.

If A and B are two angles where A + B = 90º , that is, B = 90º - A, we have:sin A = cos B, so that, sin A = cos (90º - A)cos A = sin B, so that, cos A = sin (90º - A)Similarly, tan A = cot B

Supplementary angles.Two angles are supplementary if they add up to 180 degrees.

If A and B are two angles,where A + B = 180º , that is, B = 180º - A, we have:sin A = sin B, so that, sin A = sin (180º-A)cos A = -cos B , so that, cos A = -cos (180º-A)Similarly tan A = -tan B

Using these formulas, you can calculate the trigonometric functions of an angle in the second quadrant if you know the trigonometric functions of its supplementary angle.

Angles that differ by 180ºIf A and B are two angles, that,

B – A = 180º, that is,B = 180º + A, then:sin A = -sin B, so that, sin A = -sin (180º + A)cos A = -cos B , so that, cos A = -cos (180º + A)Similarly tan A = tan B

Using these formulas, you can calculate the trigonometric functions of an angle in the third quadrant if you know the trigonometric functions of the angle that differs with it 180º. 

    Trigonometric equations are occupies six functions they are sine, cotangent, cosine, secant, tangent, and cosecant. Trigonometric equations are comprises the trigonometric function of unfamiliar angles similar equations, if all values of the unidentified angles for the functions are described and provisional equations are used only for the exacting values of the indefinite angles.

We converse about the basic trigonometric equations. Trigonometric equations are confidential as

Equations concerning more than two trigonometric functions

Equations concerning the trigonometric functions of several angles

Equations of quadratic formRemove Square RootsBy means of opposite Functions

• State all the functions into a particular trigonometric function using dissimilar trigonometric identity.

• Move the constants and numbers to the right hand side of the equation and trigonometric function to left hand side of the equation.

• Explain the trigonometric function and then get the opposite trigonometric function on both sides of the equation to calculate the unidentified angle.

Sin x = sin X1 = + k. 360

X2 = (180 - ) + k . 360Cos x = cos X = ± + k . 360

Tan x = tan x = + k . 180 k € integers

Problem 1: Solve basic trigonometic equation for x, sin x =    (0≤x≤2π)   solution:

sin x =  √ 2/ 4 = - sin 45°  

= sin (180° + 45°) = sin (360° - 45°)x = 225°, 315°

Problem 1: Solve basic trigonometric equation: cos x + √2 = - cos x

Solution:Start by rewriting the equation so that x is isolated on one side of the equation.cos x + √2 = - cos xcos x + cos x + √2 = 0cos x + cos x = -√22 cos x= -√2cos x = -√2/2

Because cos x has a period of 2π, first calculate all solutions in the interval (0, 2π), these answers are x = 3π/4 and x = 5π/4. In conclusion, add multiples of 2π to each of these answers to obtain the general form

X = 3π/4 + 2 n π and x = 5π/4 + 2 n π Where n is an integer.

• Problem 2: Solve: 4 cos2 x – 3 = 0• Solution:

Commence by rewriting the equation4 cos2 x – 3 = 04 cos2 x = 3cos2 x = `3/4`cos x = ± √3/4 = √3/2

• Because sinx has a period of 3π, first find all solutions in the interval (0, π), these solutions are x =π/6 and x= 11π/6. finally, add multiples of π to each of these solutions to get the general form

• X = π/6 + n π and x = 5π/6 + n π Where n is an integer.

• These are the examples for solving trigonometric equations.

Sin2 x + cos2 x = 1Tan2 x + 1 = sec2 xCotan2 x + 1 = cosec2 xTan Ɵ = sin Ɵ

cos ƟCotan Ɵ = 1

tan ƟCosec Ɵ = 1

sin Ɵ

Sin A = t

b t = b . Sin A Sin B = t

a t = a . Sin A

b . Sin A = a . Sin B a = b = Csin A sin Bsin C

A2 = b2 + c2 – 2 b c cos ACos A = b2 + c2 – a2

2 b c

1 . A . T2

1 b. C. Sin A2

1 a. B. Sin C2