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Futhe Mathematics
[tex](Cos {}^{4}t -Sin {}^{4}t ) \div Cos {}^{2}t[/tex]


Sagot :

Answer:

cos2t/cos²t

Step-by-step explanation:

Here the given trigonometric expression to us is ,

[tex]\longrightarrow \dfrac{cos^4t - sin^4t }{cos^2t } [/tex]

We can write the numerator as ,

[tex]\longrightarrow \dfrac{ (cos^2t)^2-(sin^2t)^2}{cos^2t } [/tex]

Recall the identity ,

[tex]\longrightarrow (a-b)(a+b)=a^2-b^2[/tex]

Using this we have ,

[tex]\longrightarrow \dfrac{(cos^2t + sin^2t)(cos^2t-sin^2t)}{cos^2t}[/tex]

Again , as we know that ,

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

Therefore we can rewrite it as ,

[tex]\longrightarrow \dfrac{1(cos^2t - sin^2t)}{cos^2t}[/tex]

Again using the first identity mentioned above ,

[tex]\longrightarrow \underline{\underline{\dfrac{(cost + sint )(cost - sint)}{cos^2t}}}[/tex]

Or else we can also write it using ,

[tex]\longrightarrow cos2\phi = cos^2\phi - sin^2\phi [/tex]

Therefore ,

[tex]\longrightarrow \underline{\underline{\dfrac{cos2t}{cos^2t}}}[/tex]

And we are done !

[tex]\rule{200}{4}[/tex]

Additional info :-

Derivation of cos²x - sin²x = cos2x :-

We can rewrite cos 2x as ,

[tex]\longrightarrow cos(x + x ) [/tex]

As we know that ,

[tex]\longrightarrow cos(y + z )= cosy.cosz - siny.sinz [/tex]

So that ,

[tex]\longrightarrow cos(x+x) = cos(x).cos(x) - sin(x)sin(x) [/tex]

On simplifying,

[tex]\longrightarrow cos(x+x) = cos^2x - sin^2x [/tex]

Hence,

[tex]\longrightarrow\underline{\underline{cos (2x) = cos^2x - sin^2x }} [/tex]

[tex]\rule{200}{4}[/tex]