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 (–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

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