Question

a question for maths lovers

try to solve it. ​

Answer 1

SoluTion :-

$\sf {1 +2sinxcosx = a}\\\\\\\sf {sin^{2}x + cos^{2}x + 2sinxcosx = a}\\\\\\\sf {(sinx + cosx)^{2} = a}\\\\\\\sf {(\frac{5}{4})^{2}=a}\\\\\\\\\sf {\therefore 32a = 32 \times (\frac{5}{4})^{2}=\frac{32 \times 25}{16}}\\\\\\\sf {\mapsto \ 32 a = 50}$

Answer 2

Step-by-step explanation:

$\bf{question \: }$

given sinx+cosx=5/4. If 1+2sinxcosx=a,then find the value of 32a.

Solution:-

We have,

$\bf \: sinx+cosx= \frac{5}{4} \\ </p><p> \\ \bf \: If \: 1+2sinxcosx=a$

. Solving the equation:-

$\bf \: sinx+cosx= \frac{5}{4} \\ \\ \bf \:$

$\bf \: squaring \: both \: sides$

$\bf { sin }^{2} + {cos}^{2} =( { \frac{5}{4} })^{2}$

GIVEN:-

$\bf \: sin^2 x+cos ^2x+2 \: sinxcosx= \frac{25}{16} \:$

$\bf \: 1+2 \: sinxcosx= \frac{25}{16} \\ \: \bf \: a = \frac{25}{16}$

$\bf1 +2sinxcosx = a \\ </p><p></p><p> \bf \: or, \\ \bf\: sin^2x + cos^2x + 2sinxcosx = a \\ </p><p></p><p> \bf \: or, \\ \bf (sinx + cosx)^2= a</p><p> \:$

Then putting the value of a from task

$\bf32 \: a = 32 × \frac{25}{16} \\ \\ \bf 32 \: a = 2 \times 25 \\ \\ </p><p></p><p>\bf \red {32 \: a = 50}</p><p></p><p></p><p> \: </p><p>$