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prove :
sin²θ + cos²θ = 1


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Sagot :

VirαtKσhliidn

Answer:

See below

Step-by-step explanation:

Here we need to prove that ,

[tex]\sf\longrightarrow sin^2\theta + cos^2\theta = 1 [/tex]

Imagine a right angled triangle with one of its acute angle as [tex]\theta[/tex] .

  • The side opposite to this angle will be perpendicular .
  • Also we know that ,

[tex]\sf\longrightarrow sin\theta =\dfrac{p}{h} \\[/tex]

[tex]\sf\longrightarrow cos\theta =\dfrac{b}{h} [/tex]

And by Pythagoras theorem ,

[tex]\sf\longrightarrow h^2 = p^2+b^2 \dots (i) [/tex]

Where the symbols have their usual meaning.

Now , taking LHS ,

[tex]\sf\longrightarrow sin^2\theta +cos^2\theta [/tex]

  • Substituting the respective values,

[tex]\sf\longrightarrow \bigg(\dfrac{p}{h}\bigg)^2+\bigg(\dfrac{b}{h}\bigg)^2\\[/tex]

[tex]\sf\longrightarrow \dfrac{p^2}{h^2}+\dfrac{b^2}{h^2}\\ [/tex]

[tex]\sf\longrightarrow \dfrac{p^2+b^2}{h^2} [/tex]

  • From equation (i) ,

[tex]\sf\longrightarrow\cancel{ \dfrac{h^2}{h^2}}\\ [/tex]

[tex]\sf\longrightarrow \bf 1 = RHS [/tex]

Since LHS = RHS ,

Hence Proved !

I hope this helps.

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