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Popular >

cos(x)-sin(x)>=-3/2

Solution

cos (x) - sin (x) \geq - \frac{3}{2}

Solution

True for all x \in R

Interval Notation

(- \infty, \infty)

Solution steps

cos (x) - sin (x) \geq - \frac{3}{2}

Use the following identity:

cos (x) - sin (x) = 2 cos (\frac{π}{4} + x)

2 cos (\frac{π}{4} + x) \geq - \frac{3}{2}

Divide both sides by

2

2 cos (\frac{π}{4} + x) \geq - \frac{3}{2}

Divide both sides by

2

\frac{2 cos ( \frac{π}{4} + x )}{2} \geq \frac{- \frac{3}{2}}{2}

Simplify

\frac{2 cos ( \frac{π}{4} + x )}{2} \geq \frac{- \frac{3}{2}}{2}

Simplify

\frac{2 cos ( \frac{π}{4} + x )}{2} : cos (\frac{π}{4} + x)

\frac{2 cos ( \frac{π}{4} + x )}{2}

Cancel the common factor:

2

= cos (\frac{π}{4} + x)

Simplify

\frac{- \frac{3}{2}}{2} : - \frac{3 2}{4}

\frac{- \frac{3}{2}}{2}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{\frac{3}{2}}{2}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

\frac{\frac{3}{2}}{2} = \frac{3}{2 2}

= - \frac{3}{2 2}

Rationalize

- \frac{3}{2 2} : - \frac{3 2}{4}

- \frac{3}{2 2}

Multiply by the conjugate

\frac{2}{2}

= - \frac{3 2}{2 2 2}

222 = 4

222

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

222 = 2 \cdot 2^{\frac{1}{2}} \cdot 2^{\frac{1}{2}} = 2^{1 + \frac{1}{2} + \frac{1}{2}}

= 2^{1 + \frac{1}{2} + \frac{1}{2}}

Add similar elements:

\frac{1}{2} + \frac{1}{2} = 2 \cdot \frac{1}{2}

= 2^{1 + 2 \cdot \frac{1}{2}}

2 \cdot \frac{1}{2} = 1

2 \cdot \frac{1}{2}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 2}{2}

Cancel the common factor:

2

= 1

= 2^{1 + 1}

Add the numbers:

1 + 1 = 2

= 2^{2}

2^{2} = 4

= 4

= - \frac{3 2}{4}

= - \frac{3 2}{4}

cos (\frac{π}{4} + x) \geq - \frac{3 2}{4}

cos (\frac{π}{4} + x) \geq - \frac{3 2}{4}

cos (\frac{π}{4} + x) \geq - \frac{3 2}{4}

Range of

cos (\frac{π}{4} + x) : - 1 \leq cos (\frac{π}{4} + x) \leq 1

Function range definition

The range of the basic

cos

function is

- 1 \leq cos (\frac{π}{4} + x) \leq 1

- 1 \leq cos (\frac{π}{4} + x) \leq 1

cos (\frac{π}{4} + x) \geq - \frac{3 2}{4} and - 1 \leq cos (\frac{π}{4} + x) \leq 1 : - 1 \leq cos (\frac{π}{4} + x) \leq 1

Let

y = cos (\frac{π}{4} + x)

Combine the intervals

y \geq - \frac{3 2}{4} and - 1 \leq y \leq 1

Merge Overlapping Intervals

y \geq - \frac{3 2}{4} and - 1 \leq y \leq 1

The intersection of two intervals is the set of numbers which are in both intervals

y \geq - \frac{3 2}{4}

and

- 1 \leq y \leq 1

- 1 \leq y \leq 1

- 1 \leq y \leq 1

True for all x

True for all x \in R

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