Question

(i) 4 + (7/8) (ii) 3*^ 5 1 5​

Answer 1

Step-by-step explanation:

Solution−

We have with us to find the value of

\red{\rm :\longmapsto\:\sf\dfrac{cos^3\beta}{cos\alpha}+\dfrac{sin^3\beta}{sin\alpha}}:⟼

cosα

cos

3

β

+

sinα

sin

3

β

can be rewritten as

\rm \: = \:\sf\dfrac{4cos^3\beta}{4cos\alpha}+\dfrac{4sin^3\beta}{4sin\alpha}=

4cosα

4cos

3

β

+

4sinα

4sin

3

β

We know,

\begin{gathered}\red{\rm :\longmapsto\:cos3x = {4cos}^{3}x - 3cosx} \\ \red{\rm :\longmapsto\:it \: means} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \red{\rm :\longmapsto\: {4cos}^{3}x = cos3x + 3cosx}\end{gathered}

:⟼cos3x=4cos

3

x−3cosx

:⟼itmeans

:⟼4cos

3

x=cos3x+3cosx

Also,

\red{\rm :\longmapsto\: {4sin}^{3}x = 3sinx - sin3x}:⟼4sin

3

x=3sinx−sin3x

So, using this, we get

\rm \: = \:\sf\dfrac{cos3 \beta + 3cos \beta }{4cos\alpha}+\dfrac{3sin \beta - sin3 \beta }{4sin\alpha}=

4cosα

cos3β+3cosβ

+

4sinα

3sinβ−sin3β

\rm \: = \:\dfrac{cos3 \beta sin \alpha + 3cos \beta sin \alpha + 3sin \beta cos \alpha - sin3 \beta cos \alpha }{4sin \alpha cos \alpha }=

4sinαcosα

cos3βsinα+3cosβsinα+3sinβcosα−sin3βcosα

\rm \: = \:\dfrac{cos3 \beta sin \alpha- sin3 \beta cos \alpha + 3(cos \beta sin \alpha + sin \beta cos \alpha)}{2(2sin \alpha cos \alpha) }

Answer 2

Given data: $4 + \frac{7}{8}$ and $3^5 X15$

To find: Simplify the given numbers

Solution:

According to the arithmetic rule we can able to solve one by one.

(i) $4 + \frac{7}{8}$

=> $\frac{32+7}{8}=\frac{39}{8}$

(ii) $3^5 X15$

=> 243 × 15 = 3645

Hence the simplification of given numbers are $\frac{39}{8}$ and 3645.