Question

sin^2 54°+cos^2 72°=3÷4

Answer 1

Cos(72) = Cos(90-18)=Sin(18) = $\frac{ \sqrt{5} - 1 }{4}$   ----(a)

Sin(54) = Sin(90-36) = Cos(36) = $\frac{ \sqrt{5} + 1 }{4}$   ----(b)

$a^{2} + b^{2}$

$\frac{5 + 1 + 2 \sqrt{5} }{16}$ + $\frac{5 + 1 - 2 \sqrt{5} }{16}$
$\frac{12}{16} = \frac{3}{4}$

Answer 2

Answer:

Cos(72) = Cos(90-18)=Sin(18) = \frac{ \sqrt{5} - 1 }{4}

4

5

−1

----(a)

Sin(54) = Sin(90-36) = Cos(36) = \frac{ \sqrt{5} + 1 }{4}

4

5

+1

----(b)

a^{2} + b^{2}a

2

+b

2

\frac{5 + 1 + 2 \sqrt{5} }{16}

16

5+1+2

5

+ \frac{5 + 1 - 2 \sqrt{5} }{16}

16

5+1−2

5

\frac{12}{16} = \frac{3}{4}

16

12

=

4

3