Question

(sin 90 + cos60 + cos45) × (sin30 - cos 0 + cos 45)

Answer 1

Answer:

Step-by-step explanation:

Don't know.. Right or wrong....

Answer 2

Answer:

Value of (sin90°+cos60°+cos45°)(sin30°-cos0°+co45°)

=$\frac{2\sqrt{2}-1}{4}$

Step-by-step explanation:

Value of (sin90°+cos60°+cos45°)(sin30°-cos0°+co45°)

=$(1+\frac{1}{2}+\frac{1}{\sqrt{2}})(\frac{1}{2}-1+\frac{1}{\sqrt{2}})$

=$(\frac{1}{2}+\frac{1}{\sqrt{2}})+1][(\frac{1}{2}+\frac{1}{\sqrt{2}})-1]$

=$(\frac{1}{2}+\frac{1}{\sqrt{2}})^{2}-1^{2}]$

/* We know the algebraic identity:

(a+b)(a-b)=a²-b² */

=$(\frac{1}{2})^{2}+(\frac{1}{\sqrt{2}})^{2}+2\times \frac{1}{2}\times \frac{1}{\sqrt{2}}-1$

= $\frac{1}{4}+\frac{1}{2}+\frac{1}{\sqrt{2}}-1$

= $\frac{1}{4}+\frac{1}{2}+\frac{\sqrt{2}}{\sqrt{2}\times \sqrt{2}}-1$

=$\frac{1}{4}+\frac{1}{2}+\frac{\sqrt{2}}{2}-1$

=$\frac{(1+2+2\sqrt{2}-4)}{4}$

=$\frac{2\sqrt{2}-1}{4}$

Therefore,.

Value of (sin90°+cos60°+cos45°)(sin30°-cos0°+co45°)

=$\frac{2\sqrt{2}-1}{4}$

•••♪