What is the value of 2tan^2(300)+3sin^2(150)-cos^2(180) ?

The value of 2tan^2(300)+3sin^2(150)-cos^2(180) is 23/4

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2tan^2(300)+3sin^2(150)-cos^2(180)

2 tan^{2} (30 0^{\circ}) + 3 sin^{2} (15 0^{\circ}) - cos^{2} (18 0^{\circ})

\frac{23}{4}

Solution steps

2 tan^{2} (30 0^{\circ}) + 3 sin^{2} (15 0^{\circ}) - cos^{2} (18 0^{\circ})

tan (30 0^{\circ}) = tan (12 0^{\circ})

= 2 tan^{2} (12 0^{\circ}) + 3 sin^{2} (15 0^{\circ}) - cos^{2} (18 0^{\circ})

Rewrite using trig identities:

tan^{2} (12 0^{\circ}) = sec^{2} (12 0^{\circ}) - 1

= 2 (sec^{2} (12 0^{\circ}) - 1) + 3 sin^{2} (15 0^{\circ}) - cos^{2} (18 0^{\circ})

Use the following trivial identity:

sec (12 0^{\circ}) = - 2

Use the following trivial identity:

sin (15 0^{\circ}) = \frac{1}{2}

Use the following trivial identity:

cos (18 0^{\circ}) = (- 1)

= 2 ((- 2)^{2} - 1) + 3 (\frac{1}{2})^{2} - (- 1)^{2}

Simplify

2 ((- 2)^{2} - 1) + 3 (\frac{1}{2})^{2} - (- 1)^{2} : \frac{23}{4}

= \frac{23}{4}

cos (2 x) = - \frac{3}{2}, - π \leq x \leq \frac{π}{2}

cos (2 θ) = cos^{2} (θ) - \frac{1}{2}

arccos (cos (\frac{7 π}{6}))

cos (6 0^{\circ})

sin^{2} (2 x) = 1

What is the value of 2tan^2(300)+3sin^2(150)-cos^2(180) ?
The value of 2tan^2(300)+3sin^2(150)-cos^2(180) is 23/4