Question

cos^8A+sin^8A=1-sin^2 2A+1/8sin^4 2A​

Answer 1

Answer:

Step-by-step explanation:

cos⁸A+sin⁸A = (Cos⁴A)² + (Sin⁴A)²

=> (Cos⁴A + Sin⁴A)² - 2Cos⁴ASin⁴A

=> [(Cos²A)²+(Sin²A)²]² - 2(CosASinA)⁴

=> [(Cos²A+Sin²A)² - 2Cos²ASin²A]² - 2(CosASinA)⁴

=> [1 - 2Cos²ASin²A]² - 2(CosASinA)⁴

=> [1 - 2(CosASinA)²]² - 2(CosASinA)⁴

//Remember: Sin2A = 2SinA CosA => SinACosA = 1/2Sin2A

=> [1 - 2(1/2Sin2A)²]²- 2(1/2Sin2A)⁴

=> [1 - 1/2Sin²2A]² - 1/8Sin⁴2A

=> 1 + 1/4Sin⁴2A - Sin²2A - 1/8Sin⁴2A

=> 1 - Sin²2A + 1/8Sin⁴2A

=> R.H.S

Answer 2

Trigonometry is the study of the relation between sides and angle of right angled triangle.

There are 6 trigonometric functions, namely

sine, cosine, tangent and their reciprocals cosecant, secant and tangent.

Clearly, we have sine and cosine in our question. Let us know more about these trigonometric functions,,

Every right angled triangle has three sides

1. Hypotenuse
2. Opposite side (Opposite to the angle A)
3. Adjacent side (Adjacent to angle A)

Sine of angle A is defined as,  $sinA=\,\,\frac{Opposite}{hypotenuse}$

Similarly, the cosine of A is defined as,  $cosA=\,\,\frac{Adjacent}{hypotenuse}$

Further, some important trigonometric proofs are:

• sin2A = 2 sinA cosA
• cos2A = $cos^2A-sin^2A$
• $sin^2A+cos^2A=1$        - - - - (1)

Given, $\bf cos^8A+sin^8A=1-sin^2 2A+\frac{sin^4 2A}{8}$

Let us take the LHS and prove it is equal to the RHS,

∴ Consider   $cos^8A+sin^8A=(cos^4A)^2+(sin^4A)^2$ - - - - (1)

We know that,   $(a+b)^2=a^2+b^2+2ab$

                $(a+b)^2-2ab=a^2+b^2+2ab-2ab\\\\(a+b)^2-2ab=a^2+b^2$ - - - (2)

∴ (1)  ⇒   $cos^8A+sin^8A=(cos^4A)^2+(sin^4A)^2$

Here $a = sin^4A$ and $b=cos^4 A$

                                      $=(cos^4A+sin^4A)^2 - 2\,cos^4Asin^4A\\\\=((cos^2A)^2+(sin^2A)^2)^2 - 2\,cos^4Asin^4A$ - - - (3)

Again, using the (2) formula for   $(cos^2A)^2+(sin^2A)^2$

Here $a= cos^2A$ and $a= sin^2A$

⇒ $(cos^2A)^2+(sin^2A)^2 = (cos^2A+sin^2A)^2-2(cosAsinA)^2$

From (a), we know that $sin^2A+cos^2A=1$

$(cos^2A)^2+(sin^2A)^2 = 1^2-2(cosAsinA)^2$

                                $= 1-2(cosAsinA)^2$ - - - -(4)

Again, from (a) we have, 2sinA cosA = sin2A

                                 ⇒  $sinA \,cosA = \frac{sin2A}{2}$  - - - - -(5)

Substituting (4) and (5) into (3) we get,

$(1-2(\frac{sin^22A}{2})^2)-2(\frac{sin^42A}{8}) = 1+\frac{sin^42A}{4}-\frac{sin^42A}{8}-sin^2A$

                                            $=1+\frac{sin^42A}{8}-sin^22A$

                                            =RHS

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