7
(cos Oº+sin 30°+sin 45°) (sin 90°+cos 60°-cos 45º )=
4 2
Answers
Answer:
Answer:
\frac{7}{4}
4
7
Explanation:
We know the values of
i) cos0° = 1
ii) sin45° = \frac{1}{\sqrt{2}}
2
1
iii) sin30° = \frac{1}{2}
2
1
iv) sin90° = 0
v) cos45° = \frac{1}{\sqrt{2}}
2
1
vi ) cos60° = \frac{1}{2}
2
1
_____________________
Now,
The value of
(cos 0+sin 45+sin 30)(sin 90-cos 45+cos 60)
= \left(1+\frac{1}{\sqrt{2}}+\frac{1}{2}\right)\left(1-\frac{1}{\sqrt{2}}+\frac{1}{2}\right)(1+
2
1
+
2
1
)(1−
2
1
+
2
1
)
= \left(\frac{3}{2}+\frac{1}{\sqrt{2}}\right)\left(\frac{3}{2}-\frac{1}{\sqrt{2}}\right)(
2
3
+
2
1
)(
2
3
−
2
1
)
= \left(\frac{3}{2}\right)^{2}-\left(\frac{1}{\sqrt{2}}\right)^{2}(
2
3
)
2
−(
2
1
)
2
/* By algebraic identity
(a-b)(a+b) = a²-b² */
= \left( \frac{9}{4}-\frac{1}{2}\right)(
4
9
−
2
1
)
= \frac{9-2}{4}
4
9−2
= \frac{7}{4}
4
7