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Trigonometric Identities
by
SBR
www.harekrishnahub.com
www.harekrishnahub.com
We have the following trigonometric identities:
𝑠𝑖𝑛2 𝜃 + 𝑐𝑜𝑠2 𝜃 = 1
1 + 𝑡𝑎𝑛2 𝜃 = 𝑠𝑒𝑐2 𝜃
1 + 𝑐𝑜𝑡2
𝜃 = 𝑐𝑜𝑠𝑒𝑐2
𝜃
Let us prove one by one.
Note:
𝑠𝑖𝑛 𝜃 2
, 𝑐𝑜𝑠 𝜃 2
, etc., are written as 𝑠𝑖𝑛2
𝜃and 𝑐𝑜𝑠2
𝜃 etc. and read as sine
squared θ
www.harekrishnahub.com
Consider a right angle triangle ABC as shown below.
Let ∠𝐴𝐵𝐶 = 90°
and ∠𝐶𝐴𝐵 = 𝜃
𝐴𝐵 = 𝑥 , 𝐵𝐶 = 𝑦 , 𝐴𝐶 = 𝑟
www.harekrishnahub.com
we have, 𝒙 𝟐
+ 𝒚 𝟐
= 𝒓 𝟐
Dividing the above equation throughout by 𝑟2
, we get
𝒙 𝟐
𝒓 𝟐 +
𝒚 𝟐
𝒓 𝟐 =
𝒓 𝟐
𝒓 𝟐
𝒙
𝒓
𝟐
+
𝒚
𝒓
𝟐
= 𝟏
But
𝑥
𝑟
= 𝑐𝑜𝑠 𝜃 and
𝑦
𝑟
= 𝑠𝑖𝑛 𝜃
∴ 𝑐𝑜𝑠 𝜃 2 + 𝑠𝑖𝑛 𝜃 2 = 1
∴ we have the identity
𝒔𝒊𝒏 𝟐
𝜽 + 𝒄𝒐𝒔 𝟐
𝜽 = 𝟏
www.harekrishnahub.com
we have, 𝒙 𝟐 + 𝒚 𝟐 = 𝒓 𝟐
Dividing the above equation throughout by 𝑥2
(for 𝒙 ≠ 𝟎) , we get
𝒙 𝟐
𝒙 𝟐
+
𝒚 𝟐
𝒙 𝟐
=
𝒓 𝟐
𝒙 𝟐
𝟏 +
𝒚
𝒙
𝟐
=
𝒓
𝒙
𝟐
But
𝑟
𝑥
= 𝑠𝑒𝑐 𝜃 and
𝑦
𝑥
= 𝑡𝑎𝑛 𝜃
∴ we have 𝟏 + 𝒕𝒂𝒏 𝜽 𝟐 = 𝒔𝒆𝒄 𝜽 𝟐
∴ we have the identity
1 + 𝒕𝒂𝒏2 𝜽 = 𝒔𝒆𝒄2 𝜽
www.harekrishnahub.com
we have, 𝒙 𝟐
+ 𝒚 𝟐
= 𝒓 𝟐
Dividing the above equation throughout by 𝑦2(for
𝒙 𝟐
𝒚 𝟐
+
𝒚 𝟐
𝒚 𝟐
=
𝒓 𝟐
𝒚 𝟐
𝒙
𝒚
𝟐
+ 𝟏 =
𝒓
𝒚
𝟐
𝒓
𝒚
= 𝒄𝒐𝒔𝒆𝒄 𝜽
𝒙
𝒚
= 𝒄𝒐𝒕 𝜽
∴ we have 𝒄𝒐𝒕 𝜽 𝟐 + 𝟏 = 𝒄𝒐𝒔𝒆𝒄 𝜽 𝟐
∴ we have the identity
1 + 𝒄𝒐𝒕2
𝜽 = 𝒄𝒐𝒔𝒆𝒄2
𝜽

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Trigonometric identities

  • 2. www.harekrishnahub.com We have the following trigonometric identities: 𝑠𝑖𝑛2 𝜃 + 𝑐𝑜𝑠2 𝜃 = 1 1 + 𝑡𝑎𝑛2 𝜃 = 𝑠𝑒𝑐2 𝜃 1 + 𝑐𝑜𝑡2 𝜃 = 𝑐𝑜𝑠𝑒𝑐2 𝜃 Let us prove one by one. Note: 𝑠𝑖𝑛 𝜃 2 , 𝑐𝑜𝑠 𝜃 2 , etc., are written as 𝑠𝑖𝑛2 𝜃and 𝑐𝑜𝑠2 𝜃 etc. and read as sine squared θ
  • 3. www.harekrishnahub.com Consider a right angle triangle ABC as shown below. Let ∠𝐴𝐵𝐶 = 90° and ∠𝐶𝐴𝐵 = 𝜃 𝐴𝐵 = 𝑥 , 𝐵𝐶 = 𝑦 , 𝐴𝐶 = 𝑟
  • 4. www.harekrishnahub.com we have, 𝒙 𝟐 + 𝒚 𝟐 = 𝒓 𝟐 Dividing the above equation throughout by 𝑟2 , we get 𝒙 𝟐 𝒓 𝟐 + 𝒚 𝟐 𝒓 𝟐 = 𝒓 𝟐 𝒓 𝟐 𝒙 𝒓 𝟐 + 𝒚 𝒓 𝟐 = 𝟏 But 𝑥 𝑟 = 𝑐𝑜𝑠 𝜃 and 𝑦 𝑟 = 𝑠𝑖𝑛 𝜃 ∴ 𝑐𝑜𝑠 𝜃 2 + 𝑠𝑖𝑛 𝜃 2 = 1 ∴ we have the identity 𝒔𝒊𝒏 𝟐 𝜽 + 𝒄𝒐𝒔 𝟐 𝜽 = 𝟏
  • 5. www.harekrishnahub.com we have, 𝒙 𝟐 + 𝒚 𝟐 = 𝒓 𝟐 Dividing the above equation throughout by 𝑥2 (for 𝒙 ≠ 𝟎) , we get 𝒙 𝟐 𝒙 𝟐 + 𝒚 𝟐 𝒙 𝟐 = 𝒓 𝟐 𝒙 𝟐 𝟏 + 𝒚 𝒙 𝟐 = 𝒓 𝒙 𝟐 But 𝑟 𝑥 = 𝑠𝑒𝑐 𝜃 and 𝑦 𝑥 = 𝑡𝑎𝑛 𝜃 ∴ we have 𝟏 + 𝒕𝒂𝒏 𝜽 𝟐 = 𝒔𝒆𝒄 𝜽 𝟐 ∴ we have the identity 1 + 𝒕𝒂𝒏2 𝜽 = 𝒔𝒆𝒄2 𝜽
  • 6. www.harekrishnahub.com we have, 𝒙 𝟐 + 𝒚 𝟐 = 𝒓 𝟐 Dividing the above equation throughout by 𝑦2(for 𝒙 𝟐 𝒚 𝟐 + 𝒚 𝟐 𝒚 𝟐 = 𝒓 𝟐 𝒚 𝟐 𝒙 𝒚 𝟐 + 𝟏 = 𝒓 𝒚 𝟐 𝒓 𝒚 = 𝒄𝒐𝒔𝒆𝒄 𝜽 𝒙 𝒚 = 𝒄𝒐𝒕 𝜽 ∴ we have 𝒄𝒐𝒕 𝜽 𝟐 + 𝟏 = 𝒄𝒐𝒔𝒆𝒄 𝜽 𝟐 ∴ we have the identity 1 + 𝒄𝒐𝒕2 𝜽 = 𝒄𝒐𝒔𝒆𝒄2 𝜽