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Prove that: ${{\sin }^{2}}2+{{\cos }^{2}}2=1$.

Last updated date: 23rd Jul 2024
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Hint: For solving this question you should know about the general form of trigonometric functions. In this problem we will solve it with the help of a right triangle and applying Pythagoras Theorem on that. That will provide us with the final answer and here we can consider the value of $\theta $ as 2.

Complete step by step answer:
According to our question it is asked of us to prove that ${{\sin }^{2}}2+{{\cos }^{2}}2=1$. Here we can see that this is given in a form of ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$. So, here we can consider the value of $\theta $ as 2. So, now our question is to prove, ${{\sin }^{2}}2+{{\cos }^{2}}2=1$. For this we will use a right triangle. And then we will apply Pythagoras Theorem for proving this and we will also keep in mind that $\dfrac{P}{H}=\sin \theta $ and $\dfrac{B}{H}=\cos \theta $. Here P stands for perpendicular, H stands for hypotenuse and B stands for base. Now if we make a right triangle as follows,
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Now, let angle A be $\theta $. We will apply the theorem of Pythagoras Theorem, that is, ${{a}^{2}}+{{b}^{2}}={{c}^{2}}$, where a is the value of perpendicular, b is the value of base and c is the value of hypotenuse. So, we will get it as,
  & \text{perpendicula}{{\text{r}}^{2}}+\text{bas}{{\text{e}}^{2}}=\text{hypotenus}{{\text{e}}^{2}} \\
 & \Rightarrow AB{{}^{2}}+B{{C}^{2}}=A{{C}^{2}} \\
Dividing the above equation by $A{{C}^{2}}$, we will get,
  & \dfrac{A{{B}^{2}}+B{{C}^{2}}}{A{{C}^{2}}}=\dfrac{A{{C}^{2}}}{A{{C}^{2}}} \\
 & \Rightarrow \dfrac{A{{B}^{2}}}{A{{C}^{2}}}+\dfrac{B{{C}^{2}}}{A{{C}^{2}}}=1 \\
We know that,
$\dfrac{P}{H}=\sin \theta $
$\dfrac{B}{H}=\cos \theta $
Now in the equation, $\dfrac{A{{B}^{2}}}{A{{C}^{2}}}+\dfrac{B{{C}^{2}}}{A{{C}^{2}}}=1$,
$\Rightarrow \dfrac{AB}{AC}=\sin \theta ,\dfrac{BC}{AC}=\cos \theta $
Thus we get the conclusion that,
${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$
And as we have assumed that $\theta =2$,
So, ${{\sin }^{2}}2+{{\cos }^{2}}2=1$.
Hence it is proved.

Note: Pythagoras Theorem, that is, $\text{perpendicula}{{\text{r}}^{2}}+\text{bas}{{\text{e}}^{2}}=\text{hypotenus}{{\text{e}}^{2}}$ is used in this question is the most basic theorem in mathematics and to be used in most questions like this with right angled triangles. Do not forget that, $\dfrac{P}{H}=\sin \theta ,\dfrac{B}{H}=\cos \theta ,\dfrac{P}{B}=\tan \theta $.