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Maths-

General

Easy

Question

Let $f open parentheses theta close parentheses equals sin invisible function application theta left parenthesis sin invisible function application theta plus sin invisible function application 3 theta right parenthesis.$ Then $f left parenthesis theta right parenthesis$ is

$p + q = 0$
$p + q = 2 n + 1$
$p + q = 2 n$
$p + q = 2 (2 n + 1)$

The correct answer is: $p plus q equals 2 open parentheses 2 n plus 1 close parentheses$

$253 blank tan invisible function application open parentheses fraction numerator p pi over denominator 4 end fraction close parentheses equals cot invisible function application open parentheses fraction numerator q pi over denominator 4 end fraction close parentheses$ if $open parentheses n element of Z close parentheses$
$rightwards double arrow fraction numerator p pi over denominator 4 end fraction equals n pi plus fraction numerator pi over denominator 2 end fraction minus fraction numerator q pi over denominator 4 end fraction$
$rightwards double arrow fraction numerator p over denominator 4 end fraction equals n plus fraction numerator 1 over denominator 2 end fraction minus fraction numerator q over denominator 4 end fraction$
$rightwards double arrow fraction numerator p plus q over denominator 4 end fraction equals fraction numerator 2 n plus 1 over denominator 2 end fraction$
$rightwards double arrow p plus q equals 2 open parentheses 2 n plus 1 close parentheses$
$equals left parenthesis sin invisible function application theta plus 3 sin invisible function application theta minus 4 sin to the power of 3 end exponent invisible function application theta right parenthesis sin invisible function application theta equals left parenthesis 4 sin invisible function application theta minus 4 sin to the power of 3 end exponent invisible function application theta right parenthesis sin invisible function application theta$
$equals 4 sin to the power of 2 end exponent invisible function application theta left parenthesis 1 minus sin to the power of 2 end exponent invisible function application theta right parenthesis equals 4 sin to the power of 2 end exponent invisible function application theta cos to the power of 2 end exponent invisible function application theta equals open parentheses 2 sin invisible function application theta cos invisible function application theta close parentheses to the power of 2 end exponent equals open parentheses sin invisible function application 2 theta close parentheses to the power of 2 end exponent greater or equal than 0$
Which is true for all $theta$

Related Questions to study

The value of $tan invisible function application 9 degree minus tan invisible function application 27 degree minus tan invisible function application 63 degree plus tan invisible function application 81 degree blank i s$

The value of $tan invisible function application 9 degree minus tan invisible function application 27 degree minus tan invisible function application 63 degree plus tan invisible function application 81 degree blank i s$

If $blank sin invisible function application x plus cos invisible function application x equals fraction numerator square root of 7 over denominator 2 end fraction blank$ where $blank x element of A comma$ then $tan invisible function application fraction numerator x over denominator 2 end fraction blank$ is equal to

If $blank sin invisible function application x plus cos invisible function application x equals fraction numerator square root of 7 over denominator 2 end fraction blank$ where $blank x element of A comma$ then $tan invisible function application fraction numerator x over denominator 2 end fraction blank$ is equal to

If $fraction numerator 3 pi over denominator 4 end fraction less than alpha less than pi comma blank$ then $square root of 2 cot invisible function application alpha plus fraction numerator 1 over denominator sin to the power of 2 end exponent invisible function application alpha end fraction end root blank$ is equal to

If $fraction numerator 3 pi over denominator 4 end fraction less than alpha less than pi comma blank$ then $square root of 2 cot invisible function application alpha plus fraction numerator 1 over denominator sin to the power of 2 end exponent invisible function application alpha end fraction end root blank$ is equal to

parallel

Hydrogen burns in air with a

Hydrogen burns in air with a

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Which is used hydrogen generators

Chemistry-General

parallel

Assertion:First ionization energy for nitrogen is lower than oxygen.
Reason :Across a period effective nuclear charge decreases.

Assertion:First ionization energy for nitrogen is lower than oxygen.
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Chemistry-General

Assertion :Outermost electronic configuration of most electropositive elements is ns² np³.
Reason :ns² np³is stable due to half filled subshell.

Assertion :Outermost electronic configuration of most electropositive elements is ns² np³.
Reason :ns² np³is stable due to half filled subshell.

Physics-General

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Physics-General

parallel

Assertion :I.E. of ₇N is more than that of ₈O as well as ₆C.
Reason : This is due to difference in reactivity towards oxygen.

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Physics-General

Assertion :Ground state configuration of Cr is 3d⁵, 4s¹.
Reason : A set of half filled orbitals containing one electron each with their spin parallel provides extra stability to the system.

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Reason : A set of half filled orbitals containing one electron each with their spin parallel provides extra stability to the system.

Physics-General

Assertion :More is the electron affinity greater is the reducing character.
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Physics-General

parallel

Assertion :Ionization potential across the period is Na < Al < Mg < Si.
Reason : Ionization potential decreases with decrease in atomic size.

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Physics-General

Assertion :Shielding effect increases as we go down the group.
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Physics-General

Assertion :Ionization enthalpy is always negative.
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Assertion :Ionization enthalpy is always negative.
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Physics-General

parallel

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