This presentation illustrates the concepts of trigonometric ratios

1. Trigonometric Ratios
by
SBR
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Consider a circle with centre 𝑶 and radius 𝒓
units. Let 𝑷 (𝒙, 𝒚) be any point on the
circumference of the circle. Join 𝑶𝑷.
Let the radius vector 𝑶𝑷 make an angle 𝜽 with
the positive 𝒙 − 𝒂𝒙𝒊𝒔.
∴ ∠𝑿𝑶𝑷 = 𝜽 and 𝑶𝑷 = 𝒓
Draw 𝑷𝑴 ⟂ the 𝒙 − 𝒂𝒙𝒊𝒔.
∴ 𝑶𝑴 = 𝒙 and 𝑷𝑴 = 𝒚

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𝑷𝑴𝑶 form a right angle triangle as shown
below.
Let us identify each of the sides of the triangle.
𝑶𝑷 = 𝒓 = 𝒙 𝟐 + 𝒚 𝟐 is the hypotenuse
𝑷𝑴 = 𝒚, the side opposite to 𝜽 is called the
opposite side.
𝑶𝑴 = 𝒙, the side adjacent to 𝜽 is called the
adjacent side.

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6 trigonometric ratios
Sine of angle θ 𝑠𝑖𝑛 𝜃
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
𝒚
𝒓
=
𝒚
𝒙 𝟐 + 𝒚 𝟐
CoSine of angle θ 𝑐𝑜𝑠 𝜃
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
𝒙
𝒓
=
𝒙
𝒙 𝟐 + 𝒚 𝟐
Tangent of angle θ 𝑡𝑎𝑛 𝜃
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒
𝒚
𝒙
for x ≠ 0
CoSecant of angle
θ
𝑐𝑜𝑠𝑒𝑐 𝜃
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒
𝒓
𝒚
=
𝒙 𝟐 + 𝒚 𝟐
𝒚
for y ≠ 0
Secant of angle θ 𝑠𝑒𝑐 𝜃
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒
𝒓
𝒙
=
𝒙 𝟐 + 𝒚 𝟐
𝒙
for x ≠ 0
CoTangent of angle
θ
𝑐𝑜𝑡 𝜃
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒
𝒙
𝒚
for y ≠ 0

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But
𝒕𝒂𝒏 𝜽 =
𝒐𝒑𝒑𝒐𝒔𝒊𝒕𝒆 𝒔𝒊𝒅𝒆
𝒂𝒅𝒋𝒂𝒄𝒆𝒏𝒕 𝒔𝒊𝒅𝒆
We have,
𝒕𝒂𝒏 𝜽 =
𝒐𝒑𝒑𝒐𝒔𝒊𝒕𝒆 𝒔𝒊𝒅𝒆
𝒉𝒚𝒑𝒐𝒕𝒆𝒏𝒖𝒔𝒆
𝒂𝒅𝒋𝒂𝒄𝒆𝒏𝒕 𝒔𝒊𝒅𝒆
𝒉𝒚𝒑𝒐𝒕𝒆𝒏𝒖𝒔𝒆
Dividing both the numerator and the denominator by hypotenuse, we get
𝒄𝒐𝒔 𝜽 =
𝒂𝒅𝒋𝒂𝒄𝒆𝒏𝒕 𝒔𝒊𝒅𝒆
𝒉𝒚𝒑𝒐𝒕𝒆𝒏𝒖𝒔𝒆
𝒔𝒊𝒏 𝜽 =
𝒐𝒑𝒑𝒐𝒔𝒊𝒕𝒆 𝒔𝒊𝒅𝒆
𝒉𝒚𝒑𝒐𝒕𝒆𝒏𝒖𝒔𝒆
∴ 𝒕𝒂𝒏 𝜽 =
𝒔𝒊𝒏 𝜽
𝒄𝒐𝒔 𝜽

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Reciprocal relations
𝒔𝒊𝒏 𝜽 =
𝑶𝒑𝒑 𝒔𝒊𝒅𝒆
𝒉𝒚𝒑
=
𝟏
𝒉𝒚𝒑
𝑶𝒑𝒑 𝒔𝒊𝒅𝒆
=
𝟏
𝒄𝒐𝒔𝒆𝒄 𝜽
𝒄𝒐𝒔 𝜽 =
𝒂𝒅𝒋 𝒔𝒊𝒅𝒆
𝒉𝒚𝒑
=
𝟏
𝒂𝒅𝒋
𝑶𝒑𝒑 𝒔𝒊𝒅𝒆
=
𝟏
𝒔𝒆𝒄 𝜽
𝒕𝒂𝒏 𝜽 =
𝑶𝒑𝒑 𝒔𝒊𝒅𝒆
𝒂𝒅𝒋 𝒔𝒊𝒅𝒆
=
𝟏
𝒂𝒅𝒋 𝒔𝒊𝒅𝒆
𝑶𝒑𝒑 𝒔𝒊𝒅𝒆
=
𝟏
𝒄𝒐𝒕 𝜽
Therefore, 𝒔𝒊𝒏 𝜽 and 𝒄𝒐𝒔𝒆𝒄 𝜽 are
reciprocal to each other
Therefore, 𝒄𝒐𝒔 𝜽 and 𝒔𝒆𝒄 𝜽 are
reciprocal to each other
Therefore, 𝒕𝒂𝒏 𝜽 and 𝒄𝒐𝒕 𝜽 are
reciprocal to each other