sin (2π + x) = sin x

cos (2π + x) = cos x

tan (2π + x) = tan x

Here x is an acute angle.

and 2π = 2 × 180° = 360°

Let’s see why there are same.

Suppose x = 390°

x = 360° + 30°

We note that

x = 390° is the same as 30°

So, we can write

cos (390°) =  cos (30°)

sin (390°) =  sin (30°)

tan (390°) =  tan (30°)

Similarly,

if the angle was

x = 750°

x = 2 × 360° + 30°

Thus, we rotate twice and come back to 30°

∴ x = 750° is the same as 30°

So, we can write

cos (750°) =  cos (30°)

sin (750°) =  sin (30°)

tan (750°) =  tan (30°)

Thus, we observe that

sin (n × 360° + x) = sin x

where n = 1, 2, 3, 4, ….

sin (n × 2π + x) = sin x

cos (n × 2π + x) = cos x

tan (n × 2π + x) = tan x

where n = 1, 2, 3, 4, ….

Note: For sec, cosec, cot … we convert them into sin, cos, tan and apply the sin, cos, tan formula

Let’s try some questions

### Find cos 390°

First, we convert 390° into radians.

We multiply by π/180

x = 390  × π/180

x = π × 390/180

x = π × 13/6

x = 13π/6

So, cos 390° = cos 13π/6

Thus,

cos 13π/6

=  cos (2π + π/6)

Since values of cos x repeats after an interval of 2π ,hence ignoring

= cos (π/6)

= cos (180/6°)

= cos 30°

= √3/2

### Find value of tan (–15 π /4)

tan (–15π/4)

As tan (–x) = – tan x

= – tan (15π/4)

= – tan (4π – π/4)

= – tan (–π/4)

As tan (–x) = – tan x

= – (– tan (π/4))

= tan (π/4)

= tan (180/4 °)

= tan (45°)

= 1

∴ tan (–15π/4) = 1

1. Chapter 3 Class 11 Trigonometric Functions
2. Concept wise

Transcript

sin (2π + x) = sin x cos (2π + x) = cos x tan (2π + x) = tan x Here x is an acute angle. and 2π = 2 × 180° = 360° Let’s see why there are same. Suppose x = 390° x = 360° + 30° We note that x = 390° is the same as 30° So, we can write cos (390°) =  cos (30°) sin (390°) =  sin (30°) tan (390°) =  tan (30°) Similarly, if the angle was x = 750° x = 2 × 360° + 30° Thus, we rotate twice and come back to 30° ∴ x = 750° is the same as 30° So, we can write cos (750°) =  cos (30°) sin (750°) =  sin (30°) tan (750°) =  tan (30°) Thus, we observe that sin (n × 360° + x) = sin x where n = 1, 2, 3, 4, …. Writing in radians sin (n × 2π + x) = sin x cos (n × 2π + x) = cos x tan (n × 2π + x) = tan x where n = 1, 2, 3, 4, …. Note: For sec, cosec, cot … we convert them into sin, cos, tan and apply the sin, cos, tan formula Let’s try some questions Find cos 390° First, we convert 390° into radians. We multiply by π/180 x = 390  × π/180 x = π × 390/180 x = π × 13/6 x = 13π/6 So, cos 390° = cos 13π/6 Thus, cos 13π/6 =  cos (2π + π/6) Since values of cos x repeats after an interval of 2π ,hence ignoring 2π = cos (π/6) = cos (180/6°) = cos 30° = √3/2 Find value of tan (–15π/4) tan (–15π/4) As tan (–x) = – tan x    = – tan (15π/4)   = – tan (4π – π/4)   = – tan (–π/4) As tan (–x) = – tan x   = – (– tan (π/4))   = tan (π/4)   = tan (180/4 °)   = tan (45°)   = 1 ∴ tan (–15π/4) = 1