sin (π/2 – x) = cos x |
cos (π/2 – x) = sin x |

sin (π/2 + x) = cos x |
cos (π/2 + x) = – sin x |

sin (3π/2 – x) = – cos x |
cos (3π/2 – x) = – sin x |

sin (3π/2 + x) = – cos x |
cos (3π/2 + x) = sin x |

sin (π – x) = sin x |
cos (π – x) = – cos x |

sin (π + x) = – sin x |
cos (π + x) = – cos x |

sin (2π – x) = – sin x |
cos (2π – x) = cos x |

sin (2π + x) = sin x |
cos (2π + x) = cos x |

### Let us learn how to find and remember these values

We follow two rules

**
1. If the angle is multiple of π/2, i.e. π/2, 3π/2, 5π/2,
**

then

sin becomes cos

cos becomes sin

If the angle is multiple of π, i.e. π, 2π, 3π,

then

sin remains sin

cos remains sin

**
2.The sign depends on the quadrant angle is in.
**

##
**
sin (
**
**
π
**
**
/2 –
**
**
x)
**

Since it is π/2,

**
sin will become cos
**

Here x is an acute angle

So, π/2 – x = 90 – x is an angle in 1st quadrant.

Since sin is positive in 1st quadrant

So, sign will be positive

∴ sin (π/2 – x) = cos x

##
**
cos (
**
**
π
**
**
/2 –
**
**
x)
**

Since it is π/2,

**
cos will become sin
**

Here x is an acute angle

So, π/2 – x = 90 – x is an angle in 1st quadrant.

Since cos is positive in 1st quadrant

So, sign will be positive

∴ cos (π/2 – x) = sin x

##
**
sin (
**
**
π
**
**
/2 +
**
**
x)
**

Since it is π/2,

**
sin will become cos
**

Here x is an acute angle

So, π/2 + x = 90 + x

90 + x is an angle which is greater than 90°, less than 180°

So, it will be in 2nd quadrant.

Since sin is positive in 2nd quadrant

So, sign will be positive

∴ sin (π/2 + x) = cos x

##
**
cos (
**
**
π
**
**
/2 +
**
**
x)
**

Since it is π/2,

**
cos will become sin
**

Here x is an acute angle

So, π/2 + x = 90 + x

90 + x is an angle which is greater than 90°, less than 180°

So, it will be in 2nd quadrant.

Since cos is negative in 2nd quadrant

So, sign will be negative

∴ cos (π/2 + x) = – sin x

##
**
sin (3
**
**
π
**
**
/2 –
**
**
x)
**

Since it is 3π/2,

**
sin will become cos
**

Here x is an acute angle

So, 3π/2 – x = 270 – x

270 – x is an angle which is greater than 180°, less than 270°

So, it will be in 3rd quadrant.

Since sin is negative in 3rd quadrant

So, sign will be negative

∴ sin (3π/2 – x) = – cos x

##
**
cos (3
**
**
π
**
**
/2 –
**
**
x)
**

Since it is 3π/2,

**
cos will become sin
**

Here x is an acute angle

So, 3π/2 – x = 270 – x

270 – x is an angle which is greater than 180°, less than 270°

So, it will be in 3rd quadrant.

Since cos is negative in 3rd quadrant

So, sign will be negative

∴ cos (3π/2 – x) = – sin x

##
**
sin (3
**
**
π
**
**
/2 +
**
**
x)
**

Since it is 3π/2,

**
sin will become cos
**

Here x is an acute angle

So, 3π/2 + x = 270 + x

270 + x is an angle which is greater than 270°, less than 360°

So, it will be in 4th quadrant.

Since sin is negative in 4th quadrant

So, sign will be negative

∴ sin (3π/2 + x) = – cos x

##
**
cos (3
**
**
π
**
**
/2 +
**
**
x)
**

Since it is 3π/2,

**
cos will become sin
**

Here x is an acute angle

So, 3π/2 + x = 270 + x

270 + x is an angle which is greater than 270°, less than 360°

So, it will be in 4th quadrant.

Since cos is positive in 4th quadrant

So, sign will be positive

∴ cos (3π/2 + x) = sin x

##
**
sin (
**
**
π
**
**
–
**
**
x)
**

Since it is π,

**
sin will remain sin
**

Here x is an acute angle

So, π – x = 180 – x

180 – x is an angle which is greater than 90°, less than 180°

So, it will be in 2nd quadrant.

Since sin is positive in 2nd quadrant

So, sign will be positive

∴ sin (π – x) = sin x

##
**
cos (
**
**
π
**
**
–
**
**
x)
**

Since it is π,

**
cos will remain cos
**

Here x is an acute angle

So, π – x = 180 – x

180 – x is an angle which is greater than 90°, less than 180°

So, it will be in 2nd quadrant.

Since cos is negative in 2nd quadrant

So, sign will be negative

∴ cos (π – x) = – cos x

##
**
sin (
**
**
π
**
**
+
**
**
x)
**

Since it is π,

**
sin will remain sin
**

Here x is an acute angle

So, π + x = 180 + x

180 + x is an angle which is greater than 180°, less than 270°

So, it will be in 3rd quadrant.

Since sin is negative in 3rd quadrant

So, sign will be negative

∴ sin (π + x) = – sin x

##
**
cos (
**
**
π
**
**
+
**
**
x)
**

Since it is π,

**
cos will remain cos
**

Here x is an acute angle

So, π + x = 180 + x

180 + x is an angle which is greater than 180°, less than 270°

So, it will be in 3rd quadrant.

Since cos is negative in 3rd quadrant

So, sign will be negative

∴ cos (π + x) = – cos x

##
**
sin (2
**
**
π
**
**
–
**
**
x)
**

Since it is 2π,

**
sin will remain sin
**

Here x is an acute angle

So, 2π – x = 360 – x

360 – x is an angle which is greater than 270°, less than 360°

So, it will be in 4th quadrant.

Since sin is negative in 4th quadrant

So, sign will be negative

∴ sin (2π – x) = – sin x

**
We can also prove it by
**

We know that sin repeats after 2π

⇒ sin (2π – x) = sin (–x)

*
Since sin (–x) = – sin x
*

∴ sin (2π – x) = – sin (x)

##
**
cos (2
**
**
π
**
**
–
**
**
x)
**

Since it is 2π,

**
cos will remain cos
**

Here x is an acute angle

So, 2π – x = 360 – x

360 – x is an angle which is greater than 270°, less than 360°

So, it will be in 4th quadrant.

Since cos is positive in 4th quadrant

So, sign will be positive

∴ cos (2π – x) = cos x

__
We can also prove it by
__

We know that cos repeats after 2π

⇒ cos (2π – x) = cos (–x)

*
Since cos (–x) = cos x
*

∴ cos (2π – x) = cos x

##
**
sin (2
**
**
π
**
**
+
**
**
x)
**

Since it is 2π,

**
sin will remain sin
**

Here x is an acute angle

So, 2π + x = 360 + x

360 – x is an angle which is greater than 360°, less than 360° + 90°

So, it will be in 1st quadrant.

Since sin is positive in 4th quadrant

So, sign will be positive

∴ sin (2π + x) = sin x

__
We can also prove it by
__

We know that sin repeats after 2π

⇒ sin (2π + x) = sin (x)

##
**
cos (2
**
**
π
**
**
+
**
**
x)
**

Since it is 2π,

**
cos will remain cos
**

Here x is an acute angle

So, 2π + x = 360 + x

360 – x is an angle which is greater than 360°, less than 360° + 90°

So, it will be in 1st quadrant.

Since cos is positive in 4th quadrant

So, sign will be positive

∴ cos (2π + x) = cos x

__
We can also prove it by
__

We know that cos repeats after 2π

⇒ cos (2π + x) = cos (x)

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